SCHOLIUM I.

In measuring boards, planks, and glafs, their sides are to be measured by a foot-rule divided into 100 equal parts; and after multiplying the sides, the decimal fractions are easily reduced to lesser denominations. The mensuration of these is easy, when they are rectangular parallelograms.

SCHOLIUM II.

If a field is to be measured, let it first be plotted on paper, by some of the methods above described, and let the figure so laid down be divided into triangles, as was shown in the preceding proposition.

The base of any triangle, or the perpendicular upon the base, or the distance of any two points of the field, is measured by applying it to the scale according to which the map is drawn.

SCHOLIUM III.

But if the field given be not in a horizontal plane, but uneven and mountainous, the scale gives the horizontal line between any two points, but not their distance measured on the uneven surface of the field. And indeed it would appear, that the horizontal plane is to be accounted the area of an uneven and hilly country. For if such ground is laid out for building on, or for planting with trees, or bearing corn, since these land perpendicular to the horizon, it is plain, that a mountainous country cannot be considered as of greater extent than those uses than the horizontal plane; nay, perhaps, for nourishing of plants, the horizontal plane may be preferable.

If, however, the area of a figure, as it lies regularly on the surface of the earth, is to be measured, this may be easily done by resolving it into triangles as it lies. The sum of their areas will be the area sought; which exceeds the area of the horizontal figure more or less, according as the field is more or less uneven.

PROPOSITION XXVIII.

Fig. 12. To find out the area of a rectangular parallelogram ABCD.—Let the side AB, for example, be 5 feet long, and BC (which constitutes with BA a right angle at B) be 17 feet. Let 17 be multiplied by 5, and the product 85 will be the number of square feet in the area of the figure ABCD. But if the parallelogram proposed is not rectangular as BEFC, its base BC multiplied into its perpendicular height AB (not into its side BE) will give its area. This is evident from art. 68. of part i.

PROPOSITION XXIX.

Fig. 13. To find the area of a given triangle.—Let the triangle BAC be given, whose base BC is supposed 9 feet long: let the perpendicular AD be drawn from the angle A opposite to the base, and let us suppose AD to be four feet. Let the half of the perpendicular be multiplied into the base, or the half of the base into the perpendicular, or take the half of the product of the whole base into the perpendicular, the product gives 18 square feet for the area of the given triangle.

But if only the sides are given, the perpendicular is found either by protracting the triangle, or by 12th and 13th 2. Eucl. or by trigonometry. But how the area of a triangle may be found from the given sides only, shall be shewn in the 31st prop.

PROPOSITION XXX.

Fig. 14. To find the area of any rectilineal figure.—If the figure be irregular, let it be resolved into triangles; and drawing perpendiculars to the bases in each of them, let the area of each triangle be found by the preceding prop. and the sum of these areas will give the area of the figure.

PROPOSITION XXXII.

Fig. 15. The area of the ordinate figure ABEFGH is equal to the product of the half circumference of the polygon, multiplied into the perpendicular drawn from the centre of the circumscribed circle to the side of the polygon.—For the ordinate figure can be resolved into as many equal triangles, as there are sides of the figure; figure; and since each triangle is equal to the product of half the base into the perpendicular, it is evident that the sum of all the triangles together, that is the polygon, is equal to the product of half the sum of the bases (that is the half of the circumference of the polygon) into the common perpendicular height of the triangles drawn from the centre C to one of the sides; for example, to AB.

**PROPOSITION XXXIII.**

Fig. 16. The area of a circle is found by multiplying the half of the periphery into the radius, or the half of the radius into the periphery.—For a circle is not different from an ordinate or regular polygon of an infinite number of sides, and the common height of the triangles into which the polygon or circle may be supposed to be divided is the radius of the circle.

Were it worth while, it were easy to demonstrate accurately this proposition, by means of the inscribed and circumscribed figures, as is done in the 5th prop. of the treatise of Archimedes concerning the dimensions of the circle.

**COROLLARY.**

Hence also it appears, that the area of the sector ABCD is produced by multiplying the half of the arc into the radius, and likewise that the area of the segment of the circle ADC is found by subtracting from the area of the sector the area of the triangle ABC.

**PROPOSITION XXXIV.**

Fig. 17. The circle is to the square of the diameter, as 11 to 14 nearly.—For if the diameter AB be supposed to be 7, the circumference AHBK will be almost 22 (by the 22d prop. of this part), and the area of the square DC will be 49; and, by the preceding prop., the area of the circle will be \(38\frac{1}{2}\); therefore the square DC will be to the inscribed circle as 49 to \(38\frac{1}{2}\), or as 98 to 77, that is, as 14 to 11. Q.E.D.

If greater exactness is required, you may proceed to any degree of accuracy: for the square DC is to the inscribed circle, as 1 to \(\frac{1}{2} + \frac{1}{3} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \frac{1}{13} + \ldots\), &c., in infinitum.

This series will be of no service for computing the area of the circle accurately, without some further artifice, because it converges at too slow a rate. The area of the circle will be found exactly enough for most purposes, by multiplying the square of the diameter by 7854, and dividing by 10,000, or cutting off four decimal places from the product; for the area of the circle is to the circumscribed square nearly as 7854 to 10,000.

**PROPOSITION XXXV.**

Fig. 18. To find the area of a given ellipse.—Let ABCD be an ellipse, whose greater diameter is BD, and the lesser AC, bisecting the greater perpendicularly in E. Let a mean proportional HF be found (by 13th 6. Eucl.) between AC and BD, and (by the 33d of this) find the area of the circle described on the diameter HF. This area is equal to the area of the ellipse ABCD. For because, as BD to AC, so the square of BD to the square of HF, (by 2. cor. 20th 6. Eucl.); but (by the 2d 12. Eucl.) as the square of BD to the square of HF, so is the circle of the diameter BD to the circle of the diameter HF; therefore as BD to AC, so is the circle of the diameter BD to the circle of the diameter HF. And (by the 5th prop. of Archimedes of spheroids) as the greater diameter BD to the lesser AC, so is the circle of the diameter BD to the ellipse ABCD. Consequently (by the 11th 5. Eucl.) the circle of the diameter BD will have the same proportion to the circle of the diameter HF, and to the ellipse ABCD. Therefore, (by 9th 5. Eucl.) the area of the circle of the diameter HF will be equal to the area of the ellipse ABCD. Q.E.D.

**SCHOLIUM.**

From this and the two preceding propositions, a method is derived of finding the area of an ellipse. There are two ways: 1st, Say, as one is to the lesser diameter, so is the greater diameter to a fourth number, (which is found by the rule of three.) Then again say, as 14 to 11, so is the 4th number found to the area sought. But the second way is shorter. Multiply the lesser diameter into the greater, and the product by 11; then divide the whole product by 14, and the quotient will be the area sought of the ellipse. For example, Let the greater diameter be 10, and the lesser 7; by multiplying 10 by 7, the product is 70; and multiplying that by 11, it is 770; and dividing 770 by 14, the quotient will be 55, which is the area of the ellipse sought.

The area of the ellipse will be found more accurately, by multiplying the product of the two diameters by 7854.

We shall add no more about other plain surfaces, whether rectilinear or curvilinear, which seldom occur in practice; but shall subjoin some propositions about measuring the surfaces of solids.

**PROPOSITION XXXVI.**

To measure the surface of any prism.—By the 14th definition of the 11th Eucl. a prism is contained by planes, of which two opposite sides (commonly called the bases) are plain rectilineal figures; which are either regular and ordinate, and measured by prop. 32. of this; or however irregular, and then they are measured by the 38th prop. The other sides are parallelograms, which are measured by prop. 28th; and the whole superficies of the prism consists of the sum of those taken altogether.

**PROPOSITION XXXVII.**

To measure the superficies of any pyramid.—Since its basis is a rectilinear figure, and the rest of the planes terminating in the top of the pyramid are triangles; these measured separately, and added together, give the surface of the pyramid required.

**PROPOSITION XXXVIII.**

To measure the superficies of any regular body.—These bodies are called regular, which are bounded by equilateral and equiangular figures. The superficies of the tetraedron consists of four equal and equiangular triangles; the superficies of the hexaedron, or cube, of six equal squares; an octaedron, of eight equal equilateral triangles; a dodecaedron, of twelve equal and ordinate pentagons; and the superficies of an icosaedron, of twenty equal and equilateral triangles. Therefore it will be easy to measure these surfaces from what has been already shown.

In the same manner we may measure the superficies of a solid contained by any planes. PROPOSITION XXXIX.

Fig. 19. To measure the superficies of a cylinder.—Because a cylinder differs very little from a prism, whose opposite planes (or bases) are ordinate figures of an infinite number of sides, it appears that the superficies of a cylinder, without the bases, is equal to an infinite number of parallelograms; the common altitude of all which is the same with the height of the cylinder, and the bases of them all differ very little from the periphery of the circle which is the base of the cylinder. Therefore this periphery multiplied into the common height, gives the superficies of the cylinder, excluding the bases; which are to be measured separately by the help of the 33d prop.

This proposition concerning the measure of the surface of the cylinder (excluding its bases) is evident from this, that when it is conceived to be spread out, it becomes a parallelogram, whose base is the periphery of the circle of the base of the cylinder stretched into a right line, and whose height is the same with the height of the cylinder.

PROPOSITION XL.

Fig. 20. To measure the surface of a right cone.—The surface of a right cone is very little different from the surface of a right pyramid, having an ordinate polygon for its base of an infinite number of sides; the surface of which (excluding the base) is equal to the sum of the triangles. The sum of the bases of these triangles is equal to the periphery of the circle of the base, and the common height of the triangles is the side of the cone AB; wherefore the sum of these triangles is equal to the product of the sum of the bases (i.e., the periphery of the base of the cone) multiplied into the half of the common height, or it is equal to the product of the periphery of the base.

If the area of the base is likewise wanted, it is to be found separately by the 33d prop. If the surface of a cone is supposed to be spread out on a plane, it will become a sector of a circle, whose radius is the side of the cone; and the arc terminating the sector is made from the periphery of the base. Whence, by corol. 33d prop. of this, its dimension may be found.

COROLLARY.

Hence it will be easy to measure the surface of a frustum of a cone cut by a plane parallel to the base.

PROPOSITION XLI.

Fig. 21. To measure the surface of a given sphere.—Let there be a sphere, whose centre is A, and let the area of its convex surface be required. Archimedes demonstrates (37. prop. 1. book of the sphere and cylinder) that its surface is equal to the area of four great circles of the sphere; that is, let the area of the great circle be multiplied by 4, and the product will give the area of the sphere; or, (by the 20th 6. and 2d 12 of Eucl.) the area of the sphere given is equal to the area of a circle whose radius is the right line BC, the diameter of the sphere. Therefore having measured (by 33d prop.) the circle described with the radius BC, this will give the surface of the sphere.

PROPOSITION XLII.

Fig. 22. To measure the surface of a segment of a sphere.—Let there be a segment cut off by the plane

ED. Archimedes demonstrates (49. and 50. 1 De sphera) that the surface of this segment, excluding the circular base, is equal to the area of a circle whose radius is the right line BE drawn from the vertex B of the segment to the periphery of the circle DE. Therefore, (by the 33d prop.) it is easily measured.

COROLLARY 1.

Hence that part of the surface of a sphere that lieth between two parallel planes is easily measured, by subtracting the surface of the lesser segment from the surface of the greater segment.

COROLLARY 2.

Hence likewise it follows, that the surface of a cylinder, described about a sphere (excluding the basis) is equal to the surface of the sphere, and the parts of the one to the parts of the other, intercepted between planes parallel to the basis of the cylinder.

Of Solid Figures and their Mensuration, comprehending likewise the Principles of gauging Vessels of all Figures.

As in the former part of this treatise we took an inch for the smallest measure in length, and an inch square for the smallest superficial measure; so now, in treating of the mensuration of solids, we take a cubical inch for the smallest solid measure. Of these 109 make a Scots pint; other liquid measures depend on this, as is generally known.

In dry measures, the firkin, by statute, contains 19½ pints; and on this depend the other dry measures: therefore, if the content of any solid be given in cubical inches, it will be easy to reduce the same to the common liquid or dry measures, and conversely to reduce these to solid inches. The liquid and dry measures, in use among other nations, are known from their writers.

"As to the English liquid measures, by act of parliament 1706, any round vessel, commonly called a cylinder, having an even bottom, being seven inches in diameter throughout, and six inches deep from the top of the inside to the bottom, (which vessel will be found by computation to contain 230 8/9 cubic inches), or any vessel containing 231 cubic inches, and no more, is deemed to be a lawful wine-gallon. An English pint therefore contains 28⅔ cubic inches; two pints make a quart; four quarts a gallon; 18 gallons a roundlet; three roundlets and an half, or 63 gallons, make a hogshead; the half of a hogshead is a barrel; one hogshead and a third, or 84 gallons, make a puncheon; one puncheon and a half, or two hogsheads, or 126 gallons, make a pipe or butt; the third part of a pipe, or 42 gallons, make a tierce; two pipes, or three puncheons, or four hogsheads, make a ton of wine. Though the English wine gallon is now fixed at 231 cubic inches, the standard kept in Guildhall being measured, before many persons of distinction, May 25th 1668, it was found to contain only 224 such inches.

"In the English beer measure, a gallon contains 282 cubic inches; consequently 35⅔ cubic inches make a pint, two pints make a quart, four quarts make a gallon." gallon, nine gallons a firkin, four firkins a barrel. In ale, eight gallons make a firkin, and 32 gallons make a barrel. By an act of the first of William and Mary, 34 gallons is the barrel, both for beer and ale, in all places, except within the weekly bills of mortality. In Scotland it is known that four gills make a mutchkin, two mutchkins make a chopin; a pint is two chopins; a quart is two pints; and a gallon is four quarts, or eight pints. The accounts of the cubi- cal inches contained in the Scots pint vary considerably from each other. According to our author, it contains 109 cubical inches. But the standard-jugs kept by the dean of guild of Edinburgh (one of which has the year 1555, with the arms of Scotland, and the town of E- dinburgh, marked upon it) having been carefully mea- sured several times, and by different persons, the Scots pint, according to those standards, was found to con- tain about 103\(\frac{4}{5}\) cubic inches. The pewterers jugs (by which the vessels in common use are made) are said to contain sometimes betwixt 105 and 106 cubic inches. A cask that was measured by the brewers of Edinburgh, before the commissioners of Excise in 1707, was found to contain 46\(\frac{3}{4}\) Scots pints; the same vessel contained 18\(\frac{1}{4}\) English ale gallons. Supposing this mensurating to be just, the Scots pint will be to the English ale- gallon as 289 to 750; and if the English ale-gallon be supposed to contain 282 cubical inches, the Scots pint will contain 108.664 cubical inches. But it is suspec- ted, on several grounds, that the experiment was not made with sufficient care and exactness.

The commissioners appointed by authority of parlia- ment to settle the measures and weights, in their act of February 19, 1618, relate, That having caused fill the Linlithgow firlot with water, they found that it con- tained 21\(\frac{1}{4}\) pints of the just Stirling jug and measure. They likewise ordain that this shall be the just and on- ly firlot; and add, That the wideness and breadth of the which firlot, under and above even over within the bounds, shall contain nineteen inches and the fifth part of an inch, and the deepness seven inches and a third part of an inch. According to this act (supposing their experiment and computation to have been accurate) the pint contained only 99.56 cubical inches; for the con- tent of such a vessel as is described in the act, is 2115.85, and this divided by 21\(\frac{1}{4}\) gives 99.56. But, by the weight of water said to fill this firlot in the same act, the measure of the pint agrees nearly with the E- dinburgh standard above mentioned.

As for the English measures of corn, the Winche- ster gallon contains 272\(\frac{4}{5}\) cubical inches; two gallons make a peck; four pecks, or eight gallons (that is 2178, cubical inches) make a bushel; and a quarter is eight bushels.

Our author says, that 19\(\frac{1}{2}\) Scots pints make a firlot. But this does not appear to be agreeable to the statute above mentioned, nor to the standard-jugs. It may be conjectured that the proportion assigned by him has been deduced from some experiment of how many pints, according to common use, were contained in the firlot. For if we suppose those pints to have been each of 108.664 cubical inches, according to the experiment

made in the 1707 before the commissioners of excise, described above; then 19\(\frac{1}{2}\) such pints will amount to 2118.04, cubical inches; which agrees nearly with 2115.85, the measure of the firlot by statute a- bove mentioned. But it is probable, that in this he followed the act 1587, where it is ordained, That the wheat-firlot shall contain 19 pints and two joucattes. A wheat firlot marked with the Linlithow stamps be- ing measured, was found to contain about 2211 cubi- cal inches. By the statute of 1618 the barley-firlot was to contain 31 pints of the just Stirling-jug. A Paris pint is 48 cubical Paris inches, and is nearly equal to an English wine-quart. The Boissian con- tains 644.86099 Paris cubical inches, or 780 36 En- glish cubical inches.

The Roman amphora was a cubical Roman tank; the congius was the eighth part of the amphora, the sexta- rium was one sixth of the congius. They divided the sextarius like the as or libra. Of dry measures, the medimnus was equal to two amphoras, that is, about \(1\frac{1}{3}\) English legal bushels; and the modius was the third part of the amphora.

**PROPOSITION XLIII.**

To find the solid content of a given prism.—By the 29th prop. let the area of the base of the prism be measured, and be multiplied by the height of the prism, the product will give the solid content of the prism.

**PROPOSITION XLIV.**

To find the solid content of a given pyramid.—The area of the base being found, (by the 30th prop.) let it be multiplied by the third part of the height of the pyra- mid, or the third part of the base by the height, the pro- duct will give the solid content, by 7th 12. Eucl.

**COROLLARY.**

If the solid content of a frustum of a pyramid is re- quired, first let the solid content of the entire pyramid be found; from which subtract the solid content of the part that is wanting, and the solid content of the broken pyramid will remain.

**PROPOSITION XLV.**

To find the content of a given cylinder.—The area of the base being found by prop. 33, if it be a circle, and by prop. 35, if it be an ellipse, (for in both cases it is a cylinder,) multiply it by the height of the cylinder, and the solid content of the cylinder will be produced.

**COROLLARY.**

Fig. 23. And in this manner may be measured the solid content of vessels and casks not much different from a cylin- der, as ABCD. If towards the middle EF it be somewhat greater, the area of the circle of the base being found (by 33d prop.) and added to the area of the middle circle EF, and the half of their sum (that is, an arithmetical mean between the area of the base and the area of the middle circle) taken for the base of the vessel, and multiplied into its height, the solid content of the given vessel will be produced.

Note, That the length of the vessel, as well as the di- ameters of the base, and of the circle EF, ought to be taken within the staves; for it is the solid content within the staves that is sought.

**PROPOSITION XLVI.**

To find the solid content of a given cone.—Let the area of the base (found by prop. 33.) be multiplied into \( \frac{1}{4} \) of the height, the product will give the solid content of the cone; for by the 10th 12. Eucl. a cone is the third part of a cylinder that has the same base and height.

**PROPOSITION XLVII.**

Fig. 24-25. To find the solid content of a frustum of a cone cut by a plane parallel to the plane of the base.—First, let the height of the entire cone be found, and thence (by the preceding prop.) its solid content; from which subtract the solid content of the cone cut off at the top, there will remain the solid content of the frustum of the cone.

How the content of the entire cone may be found, appears thus: Let ABCD be the frustum of the cone (either right or scalene, as in the figures 2. and 3.) let the cone ECD be supposed to be completed; let AG be drawn parallel to DE, and let AH and EF be perpendicular on CD; it will be (by 2d 6. Eucl.) as CG:CA::CD:CE; but (by art. 72. of part. 1.) as CA:AH::CE:EF; consequently (by 2d 5. Eucl.) as CG:AH::CD:FF; that is, as the excess of the diameter of the lesser base is to the height of the frustum, so is the diameter of the greater base to the height of the entire cone.

**COROLLARY.**

Fig. 26. Some casks whose staves are remarkably bend about the middle, and strait towards the ends may be taken for two portions of cones, without any considerable error. Thus ABEF is a frustum of a right cone, to whose base EF, on the other side, there is another similar frustum of a cone joined, EDCF. The vertices of these cones, if they be supposed to be completed, will be found at G and H. Whence, (by the preceding prop.) the solid content of such vessels may be found.

**PROPOSITION XLVIII.**

Fig. 27. A cylinder circumscribed about a sphere, that is, having its base equal to a great circle of the sphere, and its height equal to the diameter of the sphere, is to the sphere as 3 to 2.

Let ABEC be the quadrant of a circle, and ABDC the circumscribed square; and likewise the triangle ADC; by the revolution of the figure about the right line AC, as axis, a hemisphere will be generated by the quadrant, a cylinder of the same base and height by the square, and a cone by the triangle. Let these three be cut any how by the plane HF, parallel to the base AB; the section in the cylinder will be a circle whose radius is FH, in the hemisphere a circle of the radius EF, and in the cone a circle of the radius GF.

By (art. 69. of part 1.) EAq, or HFq=EFq and FAq taken together, (but AFq=FGq, because AC=CD); therefore the circle of the radius HF is equal to a circle of the radius EF together with a circle of the radius GF; and since this is true everywhere, all the circles together described by the respective radii HF (that is, the cylinder) are equal to all the circles described by the respective radii EF and FG (that is, to the hemisphere and the cone taken together); but, (by the 10th 12. Eucl.) the cone generated by the triangle DAC is one third part of the cylinder generated by the square BC. Whence it follows, that the hemisphere generated by the rotation of the quadrant ABEC is equal to the remaining two third parts of the cylinder, and that the whole sphere is \( \frac{3}{2} \) of the double cylinder circumscribed about it.

This is that celebrated 39th prop. 1. book of Archimedes of the sphere and cylinder; in which he determines the proportion of the cylinder to the sphere inscribed to be that of 3 to 2.

**COROLLARY.**

Hence it follows, that the sphere is equal to a cone whose height is equal to the semidiameter of the sphere, having for its base a circle equal to the superficies of the sphere, or to four great circles of the sphere, or to a circle whose radius is equal to the diameter of the sphere, (by prop. 41. of this.) And indeed a sphere differs very little from the sum of an infinite number of cones that have their bases in the surface of the sphere, and their common vertex in the centre of the sphere; so that the superficies of the sphere, (of whose dimension see prop. 41. of this) multiplied into the third part of the semidiameter, gives the solid content of the sphere.

**PROPOSITION XLIX.**

Fig. 28. To find the solid content of a sector of the sphere.—A spherical sector ABC (as appears by the cor. of the preceding prop.) is very little different from an infinite number of cones, having their bases in the superficies of the sphere BEC, and their common vertex in the centre. Wherefore the spherical superficies BEC being found (by prop. 42 of this), and multiplied into the third part of AB the radius of the sphere, the product will give the solid content of the sector ABC.

**COROLLARY.**

It is evident how to find the solidity of a spherical segment less than a hemisphere, by subtracting the cone ABC from the sector already found. But if the spherical segment be greater than a hemisphere, the cone corresponding must be added to the sector, to make the segment.

**PROPOSITION L.**

Fig. 29. To find the solidity of the spheroid, and of its segments cut by planes perpendicular to the axis.

In prop. 44. of this, it is shewn, that every where EH : EG :: CF : CD; but circles are as the squares described upon their rays, that is, the circle of the radius EH is to the circle of the radius EG, as CFq to CDq. And since it is so everywhere, all the circles described with the respective rays EH, (that is, the spheroid made by the rotation of the semi-ellipses AFB around the axis AB,) will be to all the circles described by the respective radii EG, (that is, the sphere described by the rotation of the semicircle ADB on the axis AB) as FCq to CDq; that is, as the spheroid to the sphere on the same axis, so is the square of the other axis of the generating ellipse to the square of the axis of the sphere.

And this holds, whether the spheroid be found by a revolution around the greater or lesser axis.

**COROLLARY 1.**

Hence it appears, that the half of the spheroid, formed by the rotation of the space AHFC around the axis AC, is double of the cone generated by the triangle AFC about the same axis; which is the 32d prop. of Archimedes of conoids and spheroids.

**COROLLARY 2.**

Hence, likewise, is evident the measure of segments of the the spheroid cut by planes perpendicular to the axis. For the segment of the spheroid made by the rotation of the space ANHE, round the axis AE, is to the segment of the sphere having the same axis AC, and made by the rotation of the segment of the circle AMGE, as CFq to CDq.

But if the measure of this solid be wanted with less labour, by the 34th prop. of Archimedes of conoids and spheroids, it will be as BE to AC+EB; so is the cone generated by the rotation of the triangle AHE round the axis AE, to the segment of the sphere made by the rotation of the space ANHE round the same axis AE; which could easily be demonstrated by the method of indivisibles.

COROLLARY 3.

Hence it is easy to find the solid content of the segment of a sphere or spheroid intercepted between two parallel planes, perpendicular to the axis. This agrees as well to the oblate as to the oblong spheroid; as is obvious.

COROLLARY 4.

Fig. 30. If a cask is to be valued as the middle piece of an oblong spheroid, cut by the two planes DC and FG, at right angles to the axis: first, let the solid content of the half spheroid ABCED be measured by the preceding prop. from which let the solidity of the segment DEC be be subtracted, and there will remain the segment ABCD; and this doubled will give the capacity of the cask required.

The following method is generally made use of for finding the solid content of such vessels. The double area of the greatest circle, that is, of that which is described by the diameter AB at the middle of the cask, is added to the area of the circle at the end, that is, of the circle DC or FG (for they are usually equal), and the third part of this sum is taken for a mean base of the cask; which therefore multiplied into the length of the cask OP, gives the content of the vessel required.

Sometimes vessels have other figures, different from those we have mentioned; the easy methods of measuring which may be learned from those who practise this art. What hath already been delivered, is sufficient for our purpose.

PROPOSITION LI.

Fig. 31. and 32. To find how much is contained in a vessel that is in part empty, whose axis is parallel to the horizon.—Let AGBH be the great circle in the middle of the cask, whose segment GBH is filled with liquor, the segment GAH being empty; the segment GBH is known, if the depth EB be known, and EH a mean proportional between the segments of the diameter AB and EB; which are found by a rod or ruler put into the vessel at the orifice. Let the basis of the cask, at a medium, be found, which suppose to be the circle CKDL; and let the segment KCL be similar to the segment GAH (which is either found by the rule of three, because as the circle AGBH is to the circle CKDL, so is the segment GAH to the segment KCL; or is found from the tables of segments made by authors); and the product of this segment multiplied by the length of the cask will give the liquid content remaining in the cask.

PROPOSITION LII.

To find the solid content of a regular and ordinate body.—A tetraedron being a pyramid, the solid content is found by the 44th prop. The hexaedron, or cube, being a kind of prism, it is measured by the 43d prop. An octaedron consists of two pyramids of the same square base and of equal heights; consequently its measure is found by the 44th prop. A dodecaedron consists of twelve pyramids having equal equilateral and equiangular pentagonal bases; and to one of these being measured (by the 44th prop. of this) and multiplied by 12, the product will be equal to the solid content of the dodecaedron. The icosaedron consists of 20 equal pyramids having triangular bases; the solid content of one of which being found (by the 44th prop.) and multiplied by 20, gives the whole solid. The bases and heights of these pyramids, if you want to proceed more exactly, may be found by trigonometry. See TRIGONOMETRY.

PROPOSITION LIII.

To find the solid content of a body however irregular.

—Let the given body be immersed into a vessel of water, having the figure of a parallelopipedon or prism, and let it be noted how much the water is raised upon the immersion of the body. For it is plain, that the space which the water fills, after the immersion of the body, exceeds the space filled before its immersion, by a space equal to the solid content of the body, however irregular. But when this excess is of the figure of a parallelopipedon or prism, it is easily measured by the 43d prop. of this, viz., by multiplying the area of the base, or mouth of the vessel, into the difference of the elevations of the water before and after immersion. Whence is found the solid content of the body given.

In the same way the solid content of a part of a body may be found, by immersing that part only in water.

There is no necessity to insist here on diminishing or enlarging solid bodies in a given proportion. It will be easy to deduce these things from the 11th and 12th books of Euclid.

"The following rules are subjoined for the ready computation of the contents of vessels, and of any solids in the measures in use in Great Britain.

"I. To find the content of a cylindric vessel in English wine gallons, the diameter of the base and altitude of the vessel being given in inches and decimals of an inch.

"Square the number of inches in the diameter of the vessel; multiply this square by the number of inches in the height; then multiply the product by the decimal fraction .0034; and this last product shall give the content in wine gallons and decimals of such a gallon. To express the rule arithmetically; let D represent the number of inches and decimals of an inch in the diameter of the vessel, and H the inches and decimals of an inch in the height of the vessel; then the content in wine-gallons shall be DDH X .0034, or DDH X .0034. Ex. Let the diameter D = 51.2 inches, the height H = 62.3 inches, then the content shall be 51.2 X 51.2 X 62.3 X .0034 = 555.27.342 wine-gallons. This rule follows from prop. 33. and 45. for, by the former, the area of the base of the vessel is in square inches DD X .7874; and by the latter, the content of the vessel in solid inches inches is DDHX.7854; which divided by 231 (the number of cubical inches in a wine-gallon) gives DDH X 0.034, the content in wine-gallons. But though the charges in the excise are made (by statute) on the supposition that the wine-gallon contains 231 cubical inches; yet it is said, that in fact, 224 cubical inches, the content of the standard measured at Guildhall (as was mentioned above) are allowed to be a wine-gallon.

II. Supposing the English ale-gallon to contain 282 cubical inches, the content of a cylindric vessel is computed in such gallons, by multiplying the square of the diameter of a vessel by its height as formerly, and their product by the decimal fraction .0.027.851: that is, the solid content in ale-gallons is DDHX.0.027.851.

III. Supposing the Scots pint to contain about 103.4 cubical inches, (which is the measure given by the standards at Edinburgh, according to experiments mentioned above), the content of a cylindric vessel is computed in Scots pints, by multiplying the square of the diameter of the vessel by its height, and the product of these by the decimal fraction .0076. Or the content of such a vessel in Scots pints is DDHX.0076.

Supposing the Winchester bushel to contain 2187 cubical inches, the content of a cylindric vessel is computed in those bushels by multiplying the square of the diameter of the vessel by the height, and the product by the decimal fraction .0.003.066. But the standard bushel having been measured by Mr Everard and others in 1696, it was found to contain only 2145.6 solid inches; and therefore it was enacted in the act for laying a duty upon malt, That every round bushel, with a plain and even bottom, being 18½ inches diameter throughout, and 8 inches deep, should be esteemed a legal Winchester bushel. According to this act (ratified in the first year of Queen Anne) the legal Winchester bushel contains only 2150.42 solid inches.

And the content of a cylindric vessel is computed in such bushels, by multiplying the square of the diameter by the height, and their product by the decimal fraction .0.003.625. Or the content of the vessel in those bushels is DDHX.0.003.625.

V. Supposing the Scots wheat firlet to contain 21½ Scots pints, (as is appointed by the statute 1618), and the pint to be conform to the Edinburgh standards above mentioned, the content of a cylindric vessel in such firlots is computed by multiplying the square of the diameter by the height, and their product by the decimal fraction .00.358. This firlot, in 1426, is appointed to contain 17 pints; in 1457, it was appointed to contain 18 pints; in 1587, it is 19½ pints; in 1628, it is 21½ pints: and though this last statute appears to have been founded on wrong computations in several respects; yet this part of the act that relates to the number of pints in the firlot seems to be the least exceptionable; and therefore we suppose the firlot to contain 21½ pints of the Edinburgh standard, or about 2167 cubical inches; which a little exceeds the Winchester bushel, from which it may have been originally copied.

VI. Supposing the bear-firlot to contain 31 Scots pints, (according to the statute 1618), and the pint

Vol. II. No. 56.

conform to the Edinburgh standards, the content of a cylindric vessel in such firlots is found by multiplying the square of the diameter by the height, and this product by .000.245.

When the section of the vessel is not a circle, but an ellipsis, the product of the greatest diameter by the least, is to be substituted in those rules for the square of the diameter.

VII. To compute the content of a vessel that may be considered as a frustum of a cone in any of those measures.

Let A represent the number of inches in the diameter of the greater base, B the number of inches in the diameter of the lesser base. Compute the square of A, the product of A multiplied by B, and the square of B, and collect these into a sum. Then find the third part of this sum, and substitute it in the preceding rules in the place of the square of the diameter; and proceed in all other respects as before. Thus, for example, the content in wine-gallons is AA X AB X BB X ¼ X H X .0034.

Or, to the square of half the sum of the diameters A and B, add one third part of the square of half their difference, and substitute this sum in the preceding rules for the square of the diameter of the vessel; for the square of A X ¼ B added to ¼ of the square of A—¼ B, gives ¼ AA X ¼ AB X ¼ BB.

VIII. When a vessel is a frustum of a parabolic conoid, measure the diameter of the section at the middle of the height of the frustum; and the content will be precisely the same as of a cylinder of this diameter, of the same height with the vessel.

IX. When a vessel is a frustum of a sphere, if you measure the diameter of the section at the middle of the height of the frustum, then compute the content of a cylinder of this diameter of the same height with the vessel, and from this subtract ¼ of the content of a cylinder of the same height, on a base whose diameter is equal to its height; the remainder will give the content of the vessel. That is, if D represent the diameter of the middle section, and H the height of the frustum, you are to substitute DD—¼ HH for the square of the diameter of the cylindric vessel in the first six rules.

X. When the vessel is a frustum of a spheroid, if the bases are equal, the content is readily found by the rule in p. 708. In other cases, let the axis of the solid be to the conjugate axis as n to 1; let D be the diameter of the middle section of the frustum, H the height or length of the frustum; and substitute in the first six rules DD—¼ HH for the square of the square of the diameter of the vessel.

XI. When the vessel is an hyperbolic conoid, let the axis of the solid be to the conjugate axis as n to 1; D the diameter of the section at the middle of the frustum, H the height or length: compute DD X ¼ HH, and substitute this sum for the square of the diameter of the cylindric vessel in the first six rules.

XII. In general, it is usual to measure any round vessel, by distinguishing it into several frustums, and taking taking the diameter of the section at the middle of each frustum; thence to compute the content of each, as if it was a cylinder of that mean diameter; and to give their sum as the content of the vessel. From the total content, computed in this manner, they subtract successively the numbers which express the circular areas that correspond to those mean diameters, each as often as there are inches in the altitude of the frustum to which it belongs, beginning with the uppermost; and in this manner calculate a table for the vessel, by which it readily appears how much liquor is at any time contained in it, by taking either the dry or wet inches; having regard to the inclination or drip of the vessel, when it has any.

This method of computing the content of a frustum from the diameter of the section at the middle of its height, is exact in that case only when it is a portion of a parabolic conoid; but in such vessels as are in common use, the error is not considerable. When the vessel is a portion of a cone or hyperbolic conoid, the content by this method is found less than the truth; but when it is a portion of a sphere or spheroid, the content computed in this manner exceeds the truth. The difference or error is always the same, in the different parts of the same or of similar vessels, when the altitude of the frustum is given. And when the altitudes are different, the error is in the triplicate ratio of the altitude. If exactness be required, the error in measuring the frustum of a conical vessel, in this manner, is \( \frac{1}{4} \) of the content of a cone similar to the vessel, of an altitude equal to the height of the frustum. In a sphere, it is \( \frac{1}{4} \) of a cylinder of a diameter and height equal to the frustum. In the spheroid and hyperbolic conoid, it is the same as in a cone generated by the right-angled triangle, contained by the two semiaxes of the figure, revolving about that side which is the semiaxis of the frustum.

In the usual method of computing a table for a vessel, by subducing from the whole content the number that expresses the uppermost area, as often as there are inches in the uppermost frustum, and afterwards the numbers for the other areas successively; it is obvious that the contents assigned by the table, when a few of the uppermost inches are dry, are stated a little too high, if the vessel stands on its lesser base, but too low when it stands on its greater base; because, when one inch is dry, for example, it is not the area at the middle of the uppermost frustum, but rather the area at the middle of the uppermost inch, that ought to be subducted from the total content, in order to find the content in this case.

XIII. To measure round timber: Let the mean circumference be found in feet and decimals of a foot; square it; multiply this square by the decimal .079577, and the product by the length. Ex. Let the mean circumference of a tree be 10 3 feet, and the length 24 feet. Then \( 10^3 \times 10^3 \times 0.79577 \times 24 = 202.615 \), is the number of cubical feet in the tree. The foundation of this rule is, that when the circumference of a circle is 1, the area is .0795774715, and that the areas of circles are as the squares of their circumferences.

But the common way used by artificers for measuring round timber, differs much from this rule. They call one fourth part of the circumference the girt, which is by them reckoned the side of a square, whose area is equal to the area of the section of the tree; therefore they square the girt, and then multiply by the length of the tree. According to their method, the tree of the last example would be computed at 159.13 cubic feet only.

How square timber is measured, will be easily understood from the preceding propositions. Fifty solid feet of hewn timber, and forty of rough timber, make a load.

XIV. To find the burden of a ship, or the number of tons it will carry, the following rule is commonly given. Multiply the length of the keel taken within board, by the breadth of the ship within board, taken from the midship beam, from plank to plank, and the product by the depth of the hold, taken from the plank below the keelson to the under part of the upper-deck plank, and divide the product by 94, the quotient is the content of the tonnage required. This rule however cannot be accurate; nor can one rule be supposed to serve for the measuring exactly the burden of ships of all sorts. Of this the reader will find more in the Memoirs of the Royal Academy of Sciences at Paris, for the year 1721.

Our author having said nothing of weights, it may be of use to add briefly, that the English Troy-pound contains 12 ounces, the ounce 20 penny-weight, and the penny-weight 24 grains; that the Averdupois pound contains 16 ounces, the ounce 16 drams, and that 112 pounds is usually called the hundred weight. It is commonly supposed, that 14 pounds Averdupois are equal to 17 pounds Troy. According to Mr Everard's experiments, 1 pound Averdupois is equal to 14 ounces 12 penny-weight and 16 grains Troy, that is, to 7000 grains; and an Averdupois ounce is 437\(\frac{1}{2}\) grains. The Scots Troy-pound (which, by the statute 1718, was to be the same with the French) is commonly supposed equal to 15\(\frac{3}{4}\) ounces English Troy, or 7560 grains. By a mean of standards kept by the Dean of Guild at Edinburgh, it is 7599\(\frac{1}{2}\) or 7600 grains. They who have measured the weights which were sent from London, after the union of the kingdoms, to be the standards by which the weights in Scotland should be made, have found the English Averdupois pound (from a medium of the several weights) to weigh 7000 grains, the same as Mr Everard; according to which, the Scots, Paris, or Amsterdam pound, will be to the pound Averdupois as 38 to 35.

The Scots Troy-stone contains 16 pounds, the pound 2 marks or 16 ounces, an ounce 16 drops, a drop 36 grains. Twenty Scots ounces make a Tron pound; but because it is usual to allow one to the score, the Tron pound is commonly 21 ounces. Sir John Skene, however, makes the Tron stone to contain only 19\(\frac{1}{2}\) pounds.