northern hemisphere, will be called the north pole of the sphere; and that which is in the middle of the southern hemisphere, the south pole.

If another great circle be drawn upon the sphere, in such a manner as to cut the equinoctial at an angle of $23\backslash frac\{1\}\{2\}$ degrees in two opposite points, it will represent the ecliptic, or circle of the sun's apparent annual motion: one half of which is on the north side of the equinoctial, and the other half on the south.

If a large stud be made to move eastward in this ecliptic, in such a manner as to go quite round it, in the time that the sphere is turned round westward 366 times upon its axis; this stud will represent the sun, changing his place every day a 365th part of the ecliptic; and going round westward, the same way as the stars do; but with a motion so much slower than the motion of the stars, that they will make 366 revolutions about the axis of the sphere, in the time that the sun makes only 365. During one half of these revolutions, the sun will be on the north side of the equinoctial; during the other half, on the south; and at the end of each half, in the equinoctial.

If we suppose the terrestrial globe in this machine to be about one inch in diameter, and the diameter of the starry sphere to be about five or six feet, a small insect on the globe would see only a very little portion of its surface; but it would see one half of the starry sphere; the convexity of the globe hiding the other half from its view. If the sphere be turned westward round the globe, and the insect could judge of the appearances which arise from that motion, it would see some stars rising to its view in the eastern side of the sphere, whilst others were setting on the western: but as all the stars are fixed to the sphere, the same stars would always rise in the same points of view on the east side, and set in the same points of view on the west side. With the sun it would be otherwise, because the sun is not fixed to any point of the sphere, but moves slowly along an oblique circle in it. And if the insect should look towards the south, and call that point of the globe, where the equinoctial in the sphere seems to cut it on the left side, the east point; and where it cuts the globe on the right side, the west point; the little animal would see the sun rise north of the east, and set north of the west, for 182$\backslash frac\{1\}\{2\}$ revolutions; after which, for as many more, the sun would rise south of the east, and set south of the west. And in the whole 365 revolutions, the sun would rise only twice in the east point, and set twice in the west. All these appearances would be the same, if the starry sphere stood still (the sun only moving in the ecliptic) and the earthly globe were turned round the axis of the sphere eastward. For, as the insect would be carried round with the globe, he would be quite insensible of its motion; and the sun and stars would appear to move westward.

We may imagine as many circles described upon the earth as we please; and we may imagine the plane of any circle described upon the earth to be continued, until it marks a circle in the concave sphere of the heavens.

The horizon is either sensible or rational. The sensible horizon is that circle which a man standing upon a large plane observes to terminate his view all around, where the heaven and earth seem to meet. The plane of our

sensible horizon continued to the heaven, divides it into two hemispheres; one visible to us, the other hid by the convexity of the earth.

The plane of the rational horizon, is supposed parallel to the plane of the sensible; to pass through the centre of the earth, and to be continued to the heavens. And although the plane of the sensible horizon touches the earth in the place of the observer, yet this plane, and that of the rational horizon, will seem to coincide in the heaven, because the whole earth is but a point compared to the sphere of the heaven.

The earth being a spherical body, the horizon, or limit of our view, must change as we change our place.

The poles of the earth, are those two points on its surface in which its axis terminates. The one is called the north pole, and the other the south pole.

The poles of the heaven, are those two points in which the earth's axis produced terminates in the heaven; so that the north pole of the heaven is directly over the north pole of the earth; and the south pole of the heaven is directly over the south pole of the earth.

The equator is a great circle upon the earth, every part of which is equally distant from either of the poles. It divides the earth into two equal parts, called the northern and southern hemispheres. If we suppose the plane of this circle to be extended to the heaven, it will mark the equinoctial therein, and will divide the heaven into two equal parts, called the northern and southern hemispheres of the heaven.

The meridian of any place is a great circle passing through that place and the poles of the earth. We may imagine as many such meridians as we please, because any place that is ever so little to the east or west of any other place, has a different meridian from that place; for no one circle can pass through any two such places and the poles of the earth.

The meridian of any place is divided by the poles into two semicircles: that which passes through the place is called the geographical, or upper meridian; and that which passes through the opposite place, is called the lower meridian.

When the rotation of the earth brings the plane of the geographical meridian to the sun, it is noon or mid-day to that place; and when the lower meridian comes to the sun, it is mid-night.

All places lying under the same geographical meridian, have their noon at the same time, and consequently all the other hours. All those places are said to have the same longitude, because no one of them lies either eastward or westward from any of the rest.

If we imagine 24 semicircles, one of which is the geographical meridian of a given place, to meet at the poles, and to divide the equator into 24 equal parts; each of these meridians will come round to the sun in 24 hours, by the earth's equable motion round its axis in that time. And, as the equator contains 360 degrees, there will be 15 degrees contained between any two of these meridians which are nearest to one another: for 24 times 15 is 360. And as the earth's motion is eastward, the sun's apparent motion will be westward, at the rate of 15 degrees each hour. Therefore,

They whose geographical meridian is 15 degrees eastward

ward from us, have noon, and every other hour, an hour sooner than we have. They whose meridian is fifteen degrees westward from us, have noon, and every other hour, an hour later than we have: and so on in proportion, reckoning one hour for every fifteen degrees.

For the ecliptic circle, signs, and degrees, see ASTRONOMY, p. 435.

The tropics are lesser circles in the heaven, parallel to the equinoctial; one on each side of it, touching the ecliptic in the points of its greatest declination; so that each tropic is $23\backslash frac\{1\}\{2\}$ degrees from the equinoctial, one on the north side of it, and the other on the south. The northern tropic touches the ecliptic at the beginning of Cancer, the southern at the beginning of Capricorn; for which reason the former is called the tropic of Cancer, and the latter the tropic of Capricorn.

The polar circles in the heaven, are each $23\backslash frac\{1\}\{2\}$ degrees from the poles, all around. That which goes round the north pole, is called the arctic circle. The south polar circle, is called the antarctic circle, from its being opposite to the arctic.

The ecliptic, tropics, and polar circles, are drawn upon the terrestrial globe, as well as upon the celestial. But the ecliptic, being a great fixed circle in the heavens, cannot properly be said to belong to the terrestrial globe; and is laid down upon it only for the convenience of solving some problems. So that, if this circle on the terrestrial globe was properly divided into the months and days of the year, it would not only suit the globe better, but would also make the problems thereon much easier.

For the earth's motion round its axis every 24 hours; its motion in the ecliptic round the sun every year; and the vicissitude of seasons; see ASTRONOMY, p. 452.

Description of the Terrestrial Globe.

[See Plate XLIV. fig. 2.]

The equator, ecliptic, and tropics, polar circles, and meridians, are laid down upon the globe in the manner already described. The ecliptic is divided into 12 signs, and each sign into 30 degrees. Each tropic is $23\backslash frac\{1\}\{2\}$ degrees from the equator, and each polar circle $23\backslash frac\{1\}\{2\}$ degrees from its respective pole. Circles are drawn parallel to the equator, at every ten degrees distance from it on each side to the poles: these circles are called parallels of latitude. On large globes there are circles drawn perpendicularly through every tenth degree of the equator, intersecting each other at the poles: but on globes of or under a foot diameter, they are only drawn through every fifteenth degree of the equator; these circles are generally called meridians, sometimes circles of longitude, and at other times hour-circles.

The globe is hung in a brass-ring, called the brazen meridian; and turns upon a wire in each pole sunk half its thickness into one side of the meridian ring; by which means, that side of the ring divides the globe into two equal parts, called the eastern and western hemispheres; as the equator divides it into two equal parts, called the northern and southern hemispheres. The ring is divided

into 360 equal parts or degrees, on the side wherein the axis of the globe turns. One half of these degrees are numbered, and reckoned, from the equator to the poles, where they end at 90: their use is to shew the latitudes of places. The degrees on the other half of the meridian ring are numbered from the poles to the equator, where they end at 90: their use is to shew how to elevate either the north or south pole above the horizon, according to the latitude of any given place, as it is north or south of the equator.

The brazen meridian is let into two notches made in a broad flat ring, called the wooden horizon; the upper surface of which divides the globe into two equal parts, called the upper and lower hemispheres. One notch is in the north point of the horizon, and the other in the south. On this horizon are several concentric circles, which contain the months and days of the year, the signs and degrees answering to the sun's place for each month and day, and the 32 points of the compass.—The graduated side of the brass meridian lies towards the east side of the horizon, and should be generally kept towards the person who works problems by the globes.

There is a small horary circle, so fixed to the north part of the brazen meridian, that the wire in the north pole of the globe is in the centre of that circle; and on the wire is an index, which goes over all the 24 hours of the circle, as the globe is turned round its axis. Sometimes there are two horary circles, one between each pole of the globe and the brazen meridian.

There is a thin slip of brass, called the quadrant of altitude, which is divided into 90 equal parts or degrees, answering exactly to so many degrees of the equator. It is occasionally fixed to the uppermost point of the brazen meridian by a nut and screw. The divisions end at the nut, and the quadrant is turned round upon it.

The Description and Use of the Armillary Sphere.

[See Plate LXXXVII. Fig. 1.]

THE exterior parts of this machine are, a compage of brass rings, which represent the principal circles of the heaven, viz. 1. The equinoctial AA, which is divided into 360 degrees (beginning at its intersection with the ecliptic in Aries) for shewing the sun's right ascension in degrees; and also into 24 hours, for shewing his right ascension in time. 2. The ecliptic BB, which is divided into 12 signs, and each sign into 30 degrees, and also into the months and days of the year; in such a manner, that the degree or point of the ecliptic in which the sun is, on any given day, stands over that day in the circle of months. 3. The tropic of Cancer CC, touching the ecliptic at the beginning of Cancer in $\backslash epsilon$, and the tropic of Capricorn DD, touching the ecliptic at the beginning of Capricorn in $\backslash sigma$; each $23\backslash frac\{1\}\{2\}$ degrees from the equinoctial circle. 4. The arctic circle E, and the antarctic circle F, each $23\backslash frac\{1\}\{2\}$ degrees from its respective pole at N and S. 5. The equinoctial colure GG, passing through the north and south poles of the heaven at N and S, and through the equinoctial points Aries and Libra, in the ecliptic. 6. The solstitial colure HH, passing through the poles of the heaven,

heaven, and through the solstitial points Cancer and Capricorn, in the ecliptic. Each quarter of the former of these colures is divided into 90 degrees, from the equinoctial to the poles of the world, for shewing the declination of the sun, moon, and stars; and each quarter of the latter, from the ecliptic at $e$ and $f$, to its poles $b$ and $d$, for shewing the latitude of the stars.

In the north pole of the ecliptic is a nut $b$, to which is fixed one end of a quadrantal wire, and to the other end a small sun $\backslash gamma$, which is carried round the ecliptic $B\; B$, by turning the nut: and in the south pole of the ecliptic is a pin $d$, on which is another quadrantal wire, with a small moon $Z$ upon it, which may be moved round by hand: but there is a particular contrivance for causing the moon to move in an orbit which crosses the ecliptic at an angle of $5\backslash frac\{1\}\{2\}$ degrees, in two opposite points called the moon's nodes; and also for shifting these points backward in the ecliptic, as the moon's nodes shift in the heaven.

Within these circular rings is a small terrestrial globe $J$, fixt on an axis $K\; K$, which extends from the north and south poles of the globe at $n$ and $s$, to those of the celestial sphere at $N$ and $S$. On this axis is fixt the flat celestial meridian $L\; L$, which may be set directly over the meridian of any place on the globe, and then turned round with the globe, so as to keep over the same meridian upon it. This flat meridian is graduated the same way as the brass meridian of a common globe, and its use is much the same. To this globe is fitted the moveable horizon $M\; M$, so as to turn upon two strong wires proceeding from its east and west points to the globe, and entering the globe at the opposite points of its equator, which is a moveable brass ring let into the globe in a groove all around its equator. The globe may be turned by hand within this ring, so as to place any given meridian upon it, directly under the celestial meridian $L\; L$. The horizon is divided into 360 degrees all around its outermost edge, within which are the points of the compass, for shewing the amplitude of the sun and moon, both in degrees and points. The celestial meridian $L\; L$ passes through two notches in the north and south points of the horizon, as in a common globe: but here, if the globe be turned round, the horizon and meridian turn with it. At the south pole of the sphere is a circle of 24 hours, fixt to the rings, and on the axis is an index which goes round that circle, if the globe be turned round its axis.

The whole fabric is supported on a pedestal $N$, and may be elevated or depressed upon the joint $O$, to any number of degrees from 0 to 90, by means of the arc $P$, which is fixed in the strong brass arm $Q$, and slides in the upright piece $R$, in which is a screw at $r$, to fix it at any proper elevation.

In the box $T$ are two wheels (as in Dr Long's sphere) and two pinions, whose axes come out at $V$ and $U$; either of which may be turned by the small winch $W$. When the winch is put upon the axis $V$, and turn backward, the terrestrial globe, with its horizon and celestial meridian, keep at rest; and the whole sphere of circles turns round from east, by south, to west, carrying the sun $\backslash gamma$, and moon $Z$, round the same way, and causing

them to rise above and set below the horizon. But when the winch is put upon the axis $U$, and turned forward, the sphere with the sun and moon keep at rest; and the earth, with its horizon and meridian, turn round from west, by south, to east; and bring the same points of the horizon to the sun and moon, to which these bodies came when the earth kept at rest, and they were carried round it; shewing that they rise and set in the same points of the horizon, and at the same times in the hour-circle, whether the motion be in the earth or in the heaven. If the earthly globe be turned, the hour-index goes round its hour-circle; but if the sphere be turned, the hour-circle goes round below the index.

And so, by this construction, the machine is equally fitted to shew either the real motion of the earth, or the apparent motion of the heaven.

To rectify the sphere for use, first slacken the screw $r$ in the upright stem $R$, and taking hold of the arm $Q$, move it up or down until the given degree of latitude for any place be at the side of the stem $R$; and then the axis of the sphere will be properly elevated, so as to stand parallel to the axis of the world, if the machine be set north and south by a small compass: this done, count the latitude from the north pole, upon the celestial meridian $L\; L$, down towards the north notch of the horizon, and set the horizon to that latitude; then, turn the nut $b$ until the sun $\backslash gamma$ comes to the given day of the year in the ecliptic, and the sun will be at its proper place for that day: find the place of the moon's ascending node, and also the place of the moon, by an Ephemeris, and set them right accordingly: lastly, turn the winch $W$, until either the sun comes to the meridian $L\; L$, or until the meridian comes to the sun (according as you want the sphere or earth to move) and set the hour-index to the XII, marked noon, and the whole machine will be rectified. — Then turn the winch, and observe when the sun or moon rise and set in the horizon, and the hour-index will shew the times thereof for the given day.

As those who understand the use of the globes will be at no loss to work many other problems by this sphere, it is needless to enlarge any farther upon it.

Directions for using Globes.

In using globes, keep the east side of the horizon towards you (unless your problem require the turning of it), which side you may know by the word East upon the horizon; for then you have the graduated side of the meridian towards you, the quadrant of altitude before you, and the globe divided exactly into two equal parts, by the graduated side of the meridian.

In working some problems, it will be necessary to turn the whole globe and horizon about, that you may look on the west side thereof; which turning will be apt to jog the ball so, as to shift away that degree of the globe which was before set to the horizon or meridian: to avoid which inconvenience, you may thrust in the feather-end of a quill between the ball of the globe and the brazen meridian; which, without hurting the ball, will keep

it from turning in the meridian, whilst you turn the west side of the horizon towards you.

PROB. I. To find the latitude and longitude of any given place upon the globe.—Turn the globe on its axis, until the given place comes exactly under that graduated side of the brazen meridian, on which the degrees are numbered from the equator; and observe what degree of the meridian the place then lies under; which is its latitude, north or south, as the place is north or south of the equator.

The globe remaining in this position, the degree of the equator, which is under the brazen meridian, is the longitude of the place which is east or west, as the place lies on the east or west side of the first meridian of the globe.—All the Atlantic Ocean, and America, is on the west side of the meridian of London; and the greatest part of Europe, and of Africa, together with all Asia, is on the east side of the meridian of London, which is reckoned the first meridian of the globe by the British geographers and astronomers.

PROB. II. The longitude and latitude of a place being given, to find that place on the globe.—Look for the given longitude in the equator (counting it eastward or westward from the first meridian, as it is mentioned to be east or west;) and bring the point of longitude in the equator to the brazen meridian, on that side which is above the south point of the horizon; then count from the equator, on the brazen meridian, to the degree of the given latitude, towards the north or south pole, according as the latitude is north or south; and under that degree of latitude on the meridian, you will have the place required.

PROB. III. To find the difference of longitude, or difference of latitude, between any two given places.—Bring each of these places to the brazen meridian, and see what its latitude is: the lesser latitude subtracted from the greater, if both places are on the same side of the equator, or both latitudes added together, if they are on different sides of it, is the difference of latitude required. And the number of degrees contained between these places, reckoned on the equator, when they are brought separately under the brazen meridian, is their difference of longitude; if it be less than 180: but if more, let it be subtracted from 360, and the remainder is the difference of longitude required. Or,

Having brought one of the places to the brazen meridian, and set the hour-index to XII, turn the globe until the other place comes to the brazen meridian, and the number of hours and parts of an hour, past over by the index, will give the longitude in time; which may be easily reduced to degrees, by allowing 15 degrees for every hour, and one degree for every four minutes.

IV B. When we speak of bringing any place to the brazen meridian, it is the graduated side of the meridian that is meant.

PROB. IV. Any place being given, to find all those places that have the same longitude or latitude with it.—Bring the given place to the brazen meridian, then all those places which lie under that side of the meridian, from pole to pole, have the same longitude with the given place. Turn the globe round its axis, and all those pla-

ces which pass under the same degree of the meridian that the given place does, have the same latitude with that place.

Since all latitudes are reckoned from the equator, and all longitudes are reckoned from the first meridian, it is evident, that the point of the equator which is cut by the first meridian, has neither latitude nor longitude.—The greatest latitude is 90 degrees, because no place is more than 90 degrees from the equator. And the greatest longitude is 180 degrees, because no place is more than 180 degrees from the first meridian.

PROB. V. To find the anteci, periæci, and antipodes, of any given place.—Bring the given place to the brazen meridian; and having found its latitude, keep the globe in that situation, and count the same number of degrees of latitude from the equator towards the contrary pole; and where the reckoning ends, you have the anteci of the given place upon the globe. Those who live at the equator have no anteci.

The globe remaining in the same position, set the hour-index to the upper XII on the horary circle, and turn the globe until the index comes to the lower XII; then, the place which lies under the meridian, in the same latitude with the given place, is the periæci required. Those who live at the poles have no periæci.

As the globe now stands (with the index at the lower XII) the antipodes of the given place will be under the same point of the brazen meridian where its anteci stood before. Every place upon the globe has its antipodes.

PROB. VI. To find the distance between any two places on the globe.—Lay the graduated edge of the quadrant of altitude over both the places, and count the number of degrees intercepted between them on the quadrant; then multiply these degrees by 60, and the product will give the distance in geographical miles; but to find the distance in miles, multiply the degrees by 69$\backslash frac\{1\}\{2\}$, and the product will be the number of miles required. Or, take the distance betwixt any two places with a pair of compasses, and apply that extent to the equator; the number of degrees, intercepted between the points of the compasses, is the distance in degrees of a great circle; which may be reduced either to geographical miles, or to English miles, as above.

PROB. VII. A place on the globe being given, and its distance from any other place, to find all the other places upon the globe which are at the same distance from the given place.—Bring the given place to the brazen meridian, and screw the quadrant of altitude to the meridian, directly over that place; then keeping the globe in that position, turn the quadrant quite round upon it, and the degree of the quadrant that touches the second place will pass over all the other places which are equally distant with it from the given place.

This is the same as if one foot of a pair of compasses was set in the given place, and the other foot extended to the second place, whose distance is known; for if the compasses be then turned round the first place as a centre, the moving foot will go over all those places which are at the same distance with the second from it.

PROB. VIII. The hour of the day at any place being given, to find all those places where it is noon at that time.

time — Bring the given place to the brazen meridian, and set the index to the given hour; this done, turn the globe until the index points to the upper XII, and then all the places that lie under the brazen meridian have noon at that time.

N. B. The upper XII always stands for noon; and when the bringing of any place to the brazen meridian is mentioned, the side of that meridian on which the degrees are reckoned from the equator is meant, unless the contrary side be mentioned.

PROB. IX. The hour of the day at any place being given, to find what o'clock it then is at any other place. — Bring the given place to the brazen meridian, and set the index to the given hour; then turn the globe, until the place where the hour is required comes to the meridian, and the index will point out the hour at that place.

PROB. X. To find the sun's place in the ecliptic, and his declination, for any given day of the year. — Look on the horizon for the given day, and right against it you have the degree of the sign in which the sun is (or his place) on that day at noon. Find the same degree of that sign in the ecliptic line upon the globe, and having brought it to the brazen meridian, observe what degree of the meridian stands over it; for that is the sun's declination, reckoned from the equator.

PROB. XI. The day of the month being given, to find all those places of the earth over which the sun will pass vertically on that day. — Find the sun's place in the ecliptic for the given day, and having brought it to the brazen meridian, observe what point of the meridian is over it; then, turning the globe round its axis, all those places which pass under that point of the meridian, are the places required; for as their latitude is equal, in degrees and parts of a degree, to the sun's declination, the sun must be directly over head to each of them at its respective noon.

PROB. XII. A place being given in the torrid zone, to find those two days of the year on which the sun shall be vertical to that place. — Bring the given place to the brazen meridian, and mark the degree of latitude that is exactly over it on the meridian; then turn the globe round its axis, and observe the two degrees of the ecliptic which pass exactly under that degree of latitude; lastly, find on the wooden horizon, the two days of the year in which the sun is in those degrees of the ecliptic, and they are the days required: for on them, and none else, the sun's declination is equal to the latitude of the given place; and consequently, he will then be vertical to it at noon.

PROB. XIII. To find all those places of the north frigid zone, where the sun begins to shine constantly without setting, on any given day, from the 21st of March to the 23d of September. — On these two days, the sun is in the equinoctial, and enlightens the globe exactly from pole to pole: therefore, as the earth turns round its axis, which terminates in the poles, every place upon it will go equally through the light and the dark, and so make equal day and night to all places of the earth. But as the sun declines from the equator, towards either pole, he will shine just as many degrees round that pole, as are equal to his declination from the equator; so that no

place within that distance of the pole will then go through any part of the dark, and consequently the sun will not set to it. Now, as the sun's declination is northward, from the 21st of March to the 23d of September, he must constantly shine round the north pole all that time; and on the day that he is in the northern tropic, he shines upon the whole north frigid zone; so that no place within the north polar circle goes through any part of the dark on that day. Therefore,

Having brought the sun's place for the given day to the brazen meridian, and found his declination (by Prob. IX.) count as many degrees on the meridian, from the north pole, as are equal to the sun's declination from the equator, and mark that degree from the pole where the reckoning ends: then, turning the globe round its axis, observe what places in the north frigid zone pass directly under that mark; for they are the places required.

The like may be done for the south frigid zone, from the 23d of September to the 21st of March, during which time the sun shines constantly on the south pole.

PROB. XIV. To find the place over which the sun is vertical at any hour of a given day. — Having found the sun's declination for the given day (by Prob. IX.) mark it with a chalk on the brazen meridian: then bring the place where you are (suppose Edinburgh) to the brazen meridian, and set the index to the given hour; which done, turn the globe on its axis, until the index points to XII at noon; and the place on the globe, which is then directly under the point of the sun's declination marked upon the meridian, has the sun that moment in the zenith, or directly over head.

PROB. XV. The day and hour of a lunar eclipse being given; to find all those places of the earth to which it will be visible. — The moon is never eclipsed but when she is full, and so directly opposite to the sun, that the earth's shadow falls upon her. Therefore, whatever place of the earth the sun is vertical to at that time, the moon must be vertical to the antipodes of that place: so that the sun will be then visible to one half of the earth, and the moon to the other.

Find the place to which the sun is vertical at the given hour (by Prob. XIV.) elevate the pole to the latitude of that place, and bring the place to the upper part of the brazen meridian, as in the former problem: then, as the sun will be visible to all those parts of the globe which are above the horizon, the moon will be visible to all those parts which are below it, at the time of her greatest obscuration.

PROB. XVI. To rectify the globe for the latitude, the zenith, and the sun's place. — Find the latitude of the place (by Prob. I.) and if the place be in the northern hemisphere, raise the north pole above the north point of the horizon, as many degrees (counted from the pole upon the brazen meridian) as are equal to the latitude of the place. If the place be in the southern hemisphere, raise the south pole above the south point of the horizon, as many degrees as are equal to the latitude. Then, turn the globe till the place comes under its latitude on the brazen meridian, and fasten the quadrant of altitude so, that the chamfered edge of its nut (which is even with the graduated edge) may be joined to the zenith,

or point of latitude. This done, bring the sun's place in the ecliptic for the given day (found by Prob X) to the graduated side of the brazen meridian, and set the hour-index to XII at noon, which is the uppermost XII on the hour-circle; and the globe will be rectified.

PROB. XVII. The latitude of any place, not exceeding $66\backslash frac\{1\}\{2\}$ degrees, and the day of the month, being given; to find the time of sun rising and setting, and consequently the length of the day and night.—Having rectified the globe for the latitude, and for the sun's place on the given day (as directed in the preceding problem) bring the sun's place in the ecliptic to the eastern side of the horizon, and the hour-index will shew the time of sun rising; then turn the globe on its axis, until the sun's place comes to the western side of the horizon, and the index will shew the time of sun-setting.

The hour of sun-setting doubled, gives the length of the day; and the hour of sun rising doubled, gives the length of the night.

PROB. XVIII. The latitude of any place, and the day of the month being given; to find when the morning twilight begins, and the evening twilight ends, at that place.—This problem is often limited: for, when the sun does not go 18 degrees below the horizon, the twilight continues the whole night; and for several nights together in summer, between 49 and $66\backslash frac\{1\}\{2\}$ degrees of latitude; and the nearer to $66\backslash frac\{1\}\{2\}$, the greater is the number of these nights. But when it does begin and end, the following method will shew the time for any given day.

Rectify the globe, and bring the sun's place in the ecliptic to the eastern side of the horizon; then mark that point of the ecliptic with a chalk which is in the western side of the horizon, it being the point opposite to the sun's place: this done, lay the quadrant of altitude over the said point, and turn the globe eastward, keeping the quadrant at the chalk mark, until it be just 18 degrees high on the quadrant; and the index will point out the time when the morning twilight begins: for the sun's place will then be 18 degrees below the eastern side of the horizon. To find the time when the evening twilight ends, bring the sun's place to the western side of the horizon, and the point opposite to it, which was marked with the chalk, will be rising in the east: then, bring the quadrant over that point, and keeping it thereon, turn the globe westward, until the said point be 18 degrees above the horizon on the quadrant, and the index will shew the time when the evening twilight ends; the sun's place being then 18 degrees below the western side of the horizon.

PROB. XIX. To find on what day of the year the sun begins to shine constantly without setting, on any given place in the north frigid zone; and how long he continues to do so.—Rectify the globe to the latitude of the place, and turn it about until some point of the ecliptic, between aries and cancer, coincides with the north point of the horizon where the brazen meridian cuts it: then find, on the wooden horizon, what day of the year the sun is in that point of the ecliptic; for that is the day on which the sun begins to shine constantly on the given place, without setting. This done, turn the globe until some point of the ecliptic, between cancer and libra, coincides

with the north point of the horizon, where the brazen meridian cuts it; and find, on the wooden horizon, on what day the sun is in that point of the ecliptic; which is the day that the sun leaves off constantly shining on the said place, and rises and sets to it as to other places on the globe. The number of natural days, or compleat revolutions of the sun about the earth, between the two days above found, is the time that the sun keeps constantly above the horizon without setting: for all that portion of the ecliptic, which lies between the two points which intersect the horizon in the very north, never sets below it; and there is just as much of the opposite part of the ecliptic that never rises; therefore, the sun will keep as long constantly below the horizon in winter, as above it in summer.

PROB. XX. To find in what latitude the sun shines constantly without setting, for any length of time less than $182\backslash frac\{1\}\{2\}$ of our days and nights.—Find a point in the ecliptic half as many degrees from the beginning of cancer (either toward aries or libra) as there are natural days in the time given; and bring that point to the north side of the brazen meridian, on which the degrees are numbered from the pole towards the equator: then, keep the globe from turning on its axis, and slide the meridian up or down, until the foresaid point of the ecliptic comes to the north point of the horizon, and then the elevation of the pole will be equal to the latitude required.

PROB. XXI. The latitude of a place, not exceeding $66\backslash frac\{1\}\{2\}$ degrees, and the day of the month being given; to find the sun's amplitude, or point of the compass on which he rises or sets.—Rectify the globe, and bring the sun's place to the eastern side of the horizon; then observe what point of the compass on the horizon stands right against the sun's place, for that is his amplitude at rising. This done, turn the globe westward, until the sun's place comes to the western side of the horizon, and it will cut the point of his amplitude at setting. Or, you may count the rising amplitude in degrees, from the east point of the horizon, to that point where the sun's place cuts it; and the setting amplitude, from the west point of the horizon, to the sun's place at setting.

PROB. XXII. The latitude, the sun's place, and his altitude, being given; to find the hour of the day, and the sun's azimuth, or number of degrees that he is distant from the meridian.—Rectify the globe, and bring the sun's place to the given height upon the quadrant of altitude; on the eastern side of the horizon, if the time be in the forenoon; or the western side, if it be in the afternoon: then the index will shew the hour; and the number of degrees in the horizon, intercepted between the quadrant of altitude and the south point, will be the sun's true azimuth at that time.

PROB. XXIII. The latitude, hour of the day, and the sun's place, being given; to find the sun's altitude and azimuth.—Rectify the globe, and turn it until the index points to the given hour; then lay the quadrant of altitude over the sun's place in the ecliptic, and the degree of the quadrant cut by the sun's place is his altitude at that time above the horizon; and the degree of the horizon cut by the quadrant is the sun's azimuth, reckoned from the south.

PROB. XXIV. The latitude, the sun's altitude, and his

his azimuth being given; to find his place in the ecliptic, the day of the month, and hour of the day, though they had all been left.—Rectify the globe for the latitude and zenith, and set the quadrant of altitude to the given azimuth in the horizon; keeping it there, turn the globe on its axis until the ecliptic cuts the quadrant in the given altitude: that point of the ecliptic which cuts the quadrant there, will be the sun's place; and the day of the month answering thereto, will be found over the like place of the sun on the wooden horizon. Keep the quadrant of altitude in that position; and, having brought the sun's place to the brazen meridian, and the hour-index to XII at noon, turn back the globe, until the sun's place cuts the quadrant of altitude again, and the index will shew the hour.

Any two points of the ecliptic, which are equidistant from the beginning of Cancer or of Capricorn, will have the same altitude and azimuth at the same hour, though the months be different; and therefore it requires some care in this problem, not to mistake both the month and the day of the month; to avoid which, observe, that from the 20th of March to the 21st of June, that part of the ecliptic which is between the beginning of Aries and beginning of Cancer is to be used: from the 21st of June to the 23d of September, between the beginning of Cancer and beginning of Libra: from the 23d of September to the 21st of December, between the beginning of Libra and the beginning of Capricorn; and from the 21st of December to the 20th of March, between the beginning of Capricorn and beginning of Aries. And as one can never be at a loss to know in what quarter of the year he takes the sun's altitude and azimuth, the above caution with regard to the quarters of the ecliptic will keep him right as to the month and day thereof.

PROB. XXV. To find the length of the longest day at any given place.—If the place be on the north side of the equator (find its latitude by Prob. I.) and elevate the north pole to that latitude; then, bring the beginning of Cancer to the brazen meridian, and set the hour-index to XII at noon. But if the given place be on the south side of the equator, elevate the south pole to its latitude, and bring the beginning of Capricorn to the brazen meridian, and the hour-index to XII. This done, turn the globe westward, until the beginning of Cancer or Capricorn (as the latitude is north or south) comes to the horizon; and the index will then point out the time of sun-setting, for it will have gone over all the afternoon hours, between mid-day and sun set; which length of time being doubled, will give the whole length of the day from sun-rising to sun-setting. For, in all latitudes, the sun rises as long before mid day, as he sets after it.

PROB. XXVI. To find in what latitude the longest day is, of any given length, less than 24 hours.—If the latitude be north, bring the beginning of Cancer to the brazen meridian, and elevate the north pole to about $66\backslash frac\{1\}\{2\}$ degrees; but if the latitude be south, bring the beginning of Capricorn to the meridian, and elevate the south pole to about $66\backslash frac\{1\}\{2\}$ degrees; because the longest day in north latitude is, when the sun is in the first point of Cancer; and in south latitude, when he is in the first point of Capricorn. Then set the hour-index to XII at

noon, and turn the globe westward, until the index points at half the number of hours given; which done, keep the globe from turning on its axis, and slide the meridian down in the notches, until the aforesaid point of the ecliptic (viz. Cancer or Capricorn) comes to the horizon; then, the elevation of the pole will be equal to the latitude required.

PROB. XXVII. The latitude of any place, not exceeding $66\backslash frac\{1\}\{2\}$ degrees, being given; to find in what climate the place is.—Find the length of the longest day at the given place, by Prob. XXV. and whatever be the number of hours whereby it exceedeth twelve, double that number, and the sum will give the climate in which the place is.

PROB. XXVIII. The latitude, and the day of the month, being given; to find the hour of the day when the sun shines.—Set the wooden horizon truly level, and the brazen meridian due north and south by a mariner's compass; then, having rectified the globe, stick a small sewing-needle into the sun's place in the ecliptic, perpendicular to that part of the surface of the globe: this done, turn the globe on its axis, until the needle comes to the brazen meridian, and set the hour-index to XII at noon; then, turn the globe on its axis, until the needle points exactly towards the sun (which it will do when it casts no shadow on the globe), and the index will shew the hour of the day.

The Use of the Celestial Globe.

HAVING done for the present with the terrestrial globe, we shall proceed to the use of the celestial; first premising, that as the equator, ecliptic, tropics, polar-circles, horizon, and brazen meridian, are exactly alike on both globes, all the former problems concerning the sun are solved the same way by both globes. The method also of rectifying the celestial globe is the same as rectifying the terrestrial.

N. B. The sun's place for any day of the year stands directly over that day on the horizon of the celestial globe, as it does on that of the terrestrial.

The latitude and longitude of the stars, or of all other celestial phenomena, are reckoned in a very different manner from the latitude and longitude of places on the earth: for all terrestrial latitudes are reckoned from the equator; and longitudes from the meridian of some remarkable place, as of London by the British, and of Paris by the French. But the astronomers of all nations agree in reckoning the latitudes of the moon, stars, planets, and comets, from the ecliptic; and their longitudes from the equinoctial colure, in that semi-circle of which cuts the ecliptic at the beginning of Aries; and thence eastward, quite round, to the same semi-circle again. Consequently those stars which lie between the equinoctial and the northern half of the ecliptic, have north declination and south latitude; those which lie between the equinoctial and the southern half of the ecliptic, have south declination and north latitude; and all those which lie between the tropics and poles, have their declinations and latitudes of the same denomination.

There are six great circles on the celestial globe, which cut

ARMILLARY

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WORLD

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A. Bell & Son

WEST

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WORLD

This is a historical map of Europe, titled 'Europe' in a large, ornate script at the top center. The map is oriented with North at the top. It includes a grid of latitude and longitude lines, with labels for degrees and minutes. A compass rose is located in the upper left quadrant, indicating North. The map is densely populated with labels for various geographical features, cities, and regions. Key regions labeled include:

• Europe: The central continent, with sub-labels for 'NORTH GERMANY', 'SOUTH GERMANY', 'POLAND', 'RUSSIA', 'TURKEY', 'ITALY', 'SPAIN', 'PORTUGAL', 'FRANCE', 'ENGLAND', 'SCOTLAND', 'IRELAND', 'SWITZERLAND', 'AUSTRIA', 'HUNGARY', 'GREAT BRITAIN', 'ISLE OF BRITAIN', 'SCOTLAND', 'IRELAND', 'SWITZERLAND', 'AUSTRIA', 'HUNGARY', 'GREAT BRITAIN', 'ISLE OF BRITAIN'.
• Asia: Labeled 'ASIA' at the bottom, with sub-labels for 'HOLLY LAND', 'SYRIA', 'PHOENICE', 'EGYPT', 'AFRICA', 'LIBERIA', 'TIBERIA', 'ARMENIA', 'MESOPOTAMIA', 'PERIA', 'INDIA', 'CHINA', 'JAPAN'.
• Africa: Labeled 'AFRICA' on the left, with sub-labels for 'NORTH AFRICA', 'SPAIN', 'PORTUGAL', 'FRANCE', 'ENGLAND', 'SCOTLAND', 'IRELAND', 'SWITZERLAND', 'AUSTRIA', 'HUNGARY', 'GREAT BRITAIN', 'ISLE OF BRITAIN'.
• Oceans: 'OCEAN' at the top, 'MEDITERRANEAN SEA' in the center, 'BLACK SEA' at the bottom, 'NORTH ATLANTIC OCEAN' on the left, 'SOUTH ATLANTIC OCEAN' on the right.
• Other Regions: 'ISLAND OF IBERIA' at the top right, 'NORTH OF AFRICA' at the bottom right, 'NORTH OF EUROPE' at the top left.

18th Century

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DE GAMAS LAND

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Meridional Line

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HYKLEBYN OGGZY
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Meridian London

Tropic of Capricorn

Longitude from London

e. d. Balle's map.

This is a historical map of the Gulf of Mexico and surrounding regions. The map is oriented with North at the top. The Gulf of Mexico is the central feature, with the Gulf of California to the west and the Gulf of St. Lawrence to the north. The map includes a compass rose in the upper right quadrant, a grid of latitude and longitude lines, and various geographical labels. The map is framed by a border with the text 'MEXICO' at the top and 'GULF OF CALIFORNIA' at the bottom. The map is oriented with North at the top.

140 m. longitude from London

SOUTH SEA

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Pointe de AFRICA

cut the ecliptic perpendicularly, and meet in two opposite points in the polar circles; which points are each ninety degrees from the ecliptic, and are called its poles. These polar points divide those circles into 12 semicircles; which cut the ecliptic at the beginnings of the 12 signs. They resemble so many meridians on the terrestrial globe; and as all places which lie under any particular meridian-semicircle on that globe, have the same longitude, so all those points of the heaven, through which any of the above semicircles are drawn, have the same longitude.—And as the greatest latitudes on the earth are at the north and south poles of the earth, so the greatest latitudes in the heaven are at the north and south poles of the ecliptic.

For the division of the stars into constellations, &c. see ASTRONOMY, p. 486.

PROB. I. To find the right ascension and declination of the sun, or any fixed star.—Bring the sun's place in the ecliptic to the brazen meridian; then that degree in the equinoctial which is cut by the meridian, is the sun's right ascension; and that degree of the meridian which is over the sun's place, is his declination. Bring any fixed star to the meridian, and its right ascension will be cut by the meridian in the equinoctial; and the degree of the meridian that stands over it, is its declination.

So that the right ascension and declination, on the celestial globe, are found in the same manner as longitude and latitude on the terrestrial.

PROB. II. To find the latitude and longitude of any star.—If the given star be on the north side of the ecliptic, place the 90th degree of the quadrant of altitude on the north pole of the ecliptic, where the twelve semicircles meet, which divide the ecliptic into the 12 signs: but if the star be on the south side of the ecliptic, place the 90th degree of the quadrant on the south pole of the ecliptic: keeping the 90th degree of the quadrant on the proper pole, turn the quadrant about, until its graduated edge cuts the star: then, the number of degrees in the quadrant, between the ecliptic and the star, is its latitude; and the degree of the ecliptic, cut by the quadrant, is the star's longitude, reckoned according to the sign in which the quadrant then is.

PROB. III. To represent the face of the starry firmament, as seen from any given place of the earth, at any hour of the night.—Rectify the celestial globe for the given latitude, the zenith, and sun's place, in every respect, as taught by the XVIth problem, for the terrestrial; and turn it about, until the index points to the given hour: then, the upper hemisphere of the globe will represent the visible half of the heaven for that time: all the stars upon the globe being then in such situations, as exactly correspond to those in the heaven. And if the globe be placed duly north and south, by means of a small sea-compass, every star in the globe will point toward the like star in the heaven: by which means, the constellations and remarkable stars may be easily known. All those stars which are in the eastern side of the horizon, are then rising in the eastern side of the heaven; all in the western, are setting in the western side; and all those under the upper part of the brazen meridian, between the south point of the horizon and the north pole, are at their greatest altitude, if the latitude of the place be north:

but if the latitude be south, those stars which lie under the upper part of the meridian, between the north point of the horizon and the south pole, are at their greatest altitude.

PROB. IV. The latitude of the place, and day of the month, being given; to find the time when any known star will rise, or be upon the meridian, or set.—Having rectified the globe, turn it about until the given star comes to the eastern side of the horizon, and the index will shew the time of the star's rising; then turn the globe westward, and when the star comes to the brazen meridian, the index will shew the time of the star's coming to the meridian of your place; lastly, turn on, until the star comes to the western side of the horizon, and the index will shew the time of the star's setting.

N. B. In northern latitudes, those stars which are less distant from the north pole, than the quantity of its elevation above the north point of the horizon, never set; and those which are less distant from the south pole, than the number of degrees by which it is depressed below the horizon, never rise: and vice versa in southern latitudes.

PROB. V. To find at what time of the year a given star will be upon the meridian, at a given hour of the night.—Bring the given star to the upper semicircle of the brazen meridian, and set the index to the given hour; then turn the globe, until the index points to XII at noon, and the upper semicircle of the meridian will then cut the sun's place, answering to the day of the year sought; which day may be easily found against the like place of the sun among the signs on the wooden horizon.

PROB. VI. The latitude, day of the month, and azimuth of any known star, being given; to find the hour of the night.—Having rectified the globe for the latitude, zenith, and sun's place, lay the quadrant of altitude to the given degree of azimuth in the horizon: then turn the globe on its axis, until the star comes to the graduated edge of the quadrant; and when it does, the index will point out the hour of the night.

PROB. VII. The latitude of the place, the day of the month, and altitude of any known star, being given; to find the hour of the night.—Rectify the globe as in the former problem, guess at the hour of the night, and turn the globe until the index points at the supposed hour; then lay the graduated edge of the quadrant of altitude over the known star, and if the degree of the star's height in the quadrant upon the globe, answers exactly to the degree of the star's observed altitude in the heaven, you have guessed exactly: but if the star on the globe is higher or lower than it was observed to be in the heaven, turn the globe backwards or forwards, keeping the edge of the quadrant upon the star, until its centre comes to the observed altitude in the quadrant; and then, the index will shew the true time of the night.

PROB. VIII. An easy method for finding the hour of the night by any two known stars, without knowing either their altitude or azimuth; and then, of finding both their altitude and azimuth, and thereby the true meridian.—Tie one end of a thread to a common musket-bullet; and, having rectified the globe as above, hold the other end of the thread in your hand, and carry it

slowly round betwixt your eye and the starry heaven, until you find it cuts any two known stars at once. Then guessing at the hour of the night, turn the globe until the index points to that time in the hour-circle; which done, lay the graduated edge of the quadrant over any one of these two stars on the globe, which the thread cut in the heaven. If the said edge of the quadrant cuts the other star also, you have guessed the time exactly; but if it does not, turn the globe slowly backwards or forwards, until the quadrant (kept upon either star) cuts them both through their centres: and then, the index will point out the exact time of the night; the degree of the horizon, cut by the quadrant, will be the true azimuth of both these stars from the south; and the stars themselves will cut their true altitude in the quadrant. At which moment, if a common azimuth-compass be so set upon a floor or level pavement, that these stars in the heaven may have the same bearing upon it (allowing for the variation of the needle) as the quadrant of altitude has in the wooden horizon of the globe, a thread extended over the north and south points of that compass will be directly in the plane of the meridian: and if a line be drawn upon the floor or pavement, along the course of the thread, and an upright wire be placed in the southmost end of the line, the shadow of the wire will fall upon that line, when the sun is on the meridian, and shines upon the pavement.

PROB. IX. To find the place of the moon, or of any planet; and thereby to shew the time of its rising, setting, and setting.—Seek in Parker's or Weaver's ephemeris the geocentric place of the moon or planet in the ecliptic, for the given day of the month; and, according to its longitude and latitude, as shewn by the ephemeris, mark the same with a chalk upon the globe. Then, ha-

ving rectified the globe, turn it round its axis westward; and as the said mark comes to the eastern side of the horizon, to the brazen meridian, and to the western side of the horizon, the index will shew at what time the planet rises, comes to the meridian, and sets, in the same manner as it would do for a fixed star.

For an explanation of the harvest moons by a globe, see ASTRONOMY, p. 463.

For the description and use of a planetary globe, see ASTRONOMY, p. 498.

For the equation of time, see ASTRONOMY, p. 458.

HAVING thus explained the use of the globes, and general principles of geography, we must refer to the maps for the situation of each particular country, with regard to longitude, latitude, &c. The use of maps is obvious from their construction. The degrees of the meridian, and parallels, shew the longitudes and latitudes of places; and the scale of miles annexed, their distances. The situation of places, with regard to each other, as well as the cardinal points, appears by inspection; the top of the map being always the north, the bottom the south, the right-hand the east, and the left the west, unless the compass usually annexed shew the contrary.

The brevity, which we are necessarily obliged to observe, prevents us from taking any notice of many particulars, which are to be found in large treatises on this subject. A general account of countries, cities, rivers, mountains, &c. is given under their respective names, as they occur in the order of the alphabet. We shall therefore conclude this article with the following table, which will serve to give an idea of the general division of the habitable earth; and at the same time serve to explain the maps in Plates 87. 88. 89. 90. 91. and 92.

The Division of the Habitable Earth, the Square Miles of each Division and Subdivision, Capital Cities, with the Distance and Bearing of each from London; also the Time of each Country compared with that of England.

4. Italy

* A degree of longitude being 4 minutes in time, therefore by having the longitude we have the time. A watch that is set to time at London would be 16 minutes too fast at Madrid, as it lies to the west of the meridian at London; and Vienna being 16 degrees and 20 minutes to the east of the meridian of London, consequently a watch set at London would be 1 hour and 5 minutes too slow at Vienna.

GEOMETRY originally signified no more than the art of measuring the earth, or any distances or dimensions within it: but at present, it denotes the science of magnitude in general; comprehending the doctrine and relations of whatever is susceptible of augmentation or diminution, considered in that light.

Hence to geometry may be referred the consideration not only of lines, surfaces, and solids; but also of time, velocity, number, weight, &c.

This science had its rise among the Egyptians, who were in a manner compelled to invent it, to remedy the confusion which generally happened in their lands, from the inundations of the river Nile, which carried away all boundaries, and effaced all the limits of their possessions. Thus this invention, which at first consisted only in measuring the lands, that every person might have what belonged to him, was called Geometry, or the art of measuring land; and it is probable that the draughts and schemes, which they were annually compelled to make, helped them to discover many excellent properties of

these figures; which speculations continued to be gradually improved, and are so to this day.

From Egypt geometry passed into Greece; where it continued to receive new improvements in the hands of Thales, Pythagoras, Archimedes, Euclid, &c. The Elements of Geometry, written by this last in fifteen books, are a most convincing proof to what perfection this science was carried among the ancients. However, it must be acknowledged, that it fell short of modern geometry; the bounds of which, what by the invention of fluxions, and the discovery of the almost infinite orders of curves, are greatly enlarged.

We may distinguish the progress of geometry into three ages; the first of which was in its meridian glory at the time when Euclid's Elements appeared; the second, beginning with Archimedes, reaches to the time of Des Cartes, who, by applying algebra to the elements of geometry, gave a new turn to this science, which has been carried to its utmost perfection by Sir Isaac Newton and Mr Leibnitz.

In treating this useful subject, we shall divide it into two parts; the first containing the general principles; and

the second, the application of these principles to the mensuration of surfaces, solids, &c.

Art. 1. A point is that which is not made up of parts, or which is of itself indivisible.

2. A line is a length without breadth, as B—

3. The extremities of a line are points; as the extremities of the line AB, are the points A and B, fig. 1. Plate XCIII.

4. If the line AB be the nearest distance between its extremes A and B, then it is called a strait line, as AB; but if it be not the nearest distance, then it is called a curve line, as AB; fig. 1.

5. A surface is that which is considered as having only length and breadth, but no thickness, as B, fig. 2.

6. The terms or boundaries of a surface are lines.

7. A plain surface is that which lies equally between its extremes.

8. The inclination between two lines meeting one another (provided they do not make one continued line,) or the opening between them, is called an angle; thus the inclination of the line AB to the line CB (fig. 3.) meeting one another at B, or the opening between the two lines AB and CB, is called an angle.

9. When the lines forming the angle are right lines, then it is called a right-lined angle, as A, fig. 4. if one of them be right and the other curved, it is called a mixed angle, as B, fig. 5. if both if them be curved, it is called a curve-lined angle, as C, fig. 6.

10. If a right line AB fall upon another DC, (fig. 7.) so as to incline neither to one side nor to the other; but make the angles ABD, ABC, on each side equal to one another; then the line AB is said to be perpendicular to the line DC, and the two angles are called right-angles.

11. An obtuse angle is that which is greater than a right one, as A, fig. 8; and an acute angle, that which is less than a right one, as B, fig. 9.

12. If a right line DC be fastened at one of its ends C, and the other end D be carried quite round, then the space comprehended is called a circle; the curve line described by the point D, is called the periphery or circumference of the circle; the fixed point C is called the centre of it. Fig. 10.

13. The describing line CD is called the radius, viz. any line drawn from the centre to the circumference; whence all radii of the same or equal circles are equal.

14. Any line drawn through the centre, and terminated both ways by the circumference, is called a diameter, as BD is a diameter of the circle BADE. And the diameter divides the circle and circumference into two equal parts, and is double the radius.

15. The circumference of every circle is supposed to be divided into 360 equal parts, called degrees; and each degree is divided into 60 equal parts, called minutes; and each minute into 60 equal parts called, seconds; and

these into thirds, fourths, &c. these parts being greater or less according as the radius is.

16. Any part of the circumference is called an arch, or arc; and is called an arc of as many degrees as it contains parts of the 360, into which the circumference was divided: Thus if AD be the $\backslash frac\{1\}\{4\}$ of the circumference, then the arc AD is an arc of 45 degrees.

17. A line drawn from one end of an arc to the other, is called a chord, and is the measure of the arc; thus the right line AB is the chord of the arc ADB, fig. 11.

18. Any part of a circle cut off by a chord, is called a segment; thus the space comprehended between the chord AB and circumference ADB (which is cut off by the chord AB) is called a segment. Whence it is plain,

1st, That all chords divide the circle into two segments.

2dly, The less the chord is, the more unequal are the segments, and e contra.

3dly, When the chord is greatest, viz. when it is a diameter, then the segments are equal, viz. each a semicircle.

19. Any part of a circle (less than a semicircle) contained between two radii and an arc, is called a sector; thus the space contained between the two radii, AC, BC, and the arc AB, is called the sector; fig. 12.

20. The right line of any arc, is a line drawn perpendicular from one end of the arc, to a diameter drawn through the other end of the same arc; thus (fig. 13.) AD is the right line of the arc AB, it being a line drawn from A, the one end of the arc AB, perpendicular to CB, a diameter passing through B, the other end of the arc AB.

Now the fines standing on the same diameter, still increase till they come to the centre, and then becoming the radius, it is plain that the radius EC is the greatest possible fine, and for that reason it is called the whole fine.

Since the whole fine EC must be perpendicular to the a diameter FB (by def. 20.) therefore producing the diameter EG, the two diameters FB, EG, must cross one another at right angles, and so the circumference of the circle must be divided by them into four parts EB, BG, GF, and FE, and these four parts are equal to one another (by def. 10.) and so EB a quadrant, or fourth part of the circumference; therefore the radius EC is always the fine of the quadrant, or fourth part of the circle EB.

Sines are said to be of so many degrees, as the arc contains parts of the 360, into which the circumference is supposed to be divided; so the radius being the fine of a quadrant, or fourth part of the circumference, which contains 90 degrees (the fourth part of 360), therefore the radius must be the fine of 90 degrees.

21. The part of the radius comprehended between the extremity of the right line and the lower end of the arc, viz. DB, is called the versed sine of the arc AB.

22. If to any point in the circumference, viz. B, there be drawn a diameter FCB, and from the point B, perpendicular to that diameter, there be drawn the line BH; that line is called a tangent to the circle in the point B; which tangent can touch the circle only in one point B, else if it touched it in more, it would go within it, and so not be a tangent but a chord, (by art. 17.)

23. The tangent of any arc AB, is a right line drawn perpendicular to a diameter through the one end of the arc B, and terminated by a line CAH, drawn from the centre through the other end A; thus BH is the tangent of the arc AB.

24. And the line which terminates the tangent, viz. CH, is called the secant of the arc AB.

25. What an arc wants of a quadrant is called the complement of that arc; thus AE, being what the arc AB wants of the quadrant EB, is called the complement of the arc AB.

26. And what an arc wants of a semicircle is called the supplement of that arc; thus since AF is what the arc AB wants of the semicircle BAF, it is the supplement of the arc AB.

27. The sine, tangent, &c. of the complement of any arc, is called the co-sine, co-tangent, &c. of that arc; thus the sine, tangent, &c. of the arc AE is called the co-sine, co-tangent, &c. of the arc AB.

28. The sine of the supplement of an arc is the same with the sine of the arc itself; for, drawing them according to the definitions, there results the self-same line.

29. A right-lined angle is measured by an arc of a circle described upon the angular point as a centre, comprehended between the two legs that form the angle; thus (fig. 14.) the angle ABD is measured by the arc AD of the circle CADE that is described upon the point B as a centre; and the angle is said to be of as many degrees as the arc is; so if the arc AD be 45 degrees, then the angle ABD is said to be an angle of 45 degrees.

Hence the angles are greater or less, according as the arc described about the angular point and terminated by the two legs contain a greater or a less number of degrees.

30. When one line falls perpendicularly on another, as AB on CD, fig. 15. then the angles are right (by the 10th def.); and describing a circle on the centre B, since the angles ABC ABD are equal, their measures must be so too, i. e. the arcs AC AD must be equal; but the whole CAD is a semicircle, since CD, a line passing through the centre B, is a diameter; therefore each of the parts AC AD is a quadrant, i. e. 90 degrees; so the measure of a right angle is always 90 degrees.

31. If one line AB fall any way upon another, CD, then the sum of the two angles ABC ABD is always equal to the sum of two right angles; fig. 16. For on the point B, describing the circle CAD, it is plain, that CAD is a semicircle (by the 14th); but CAD is equal to CA and AD the measure of the two angles; therefore the sum of the two angles is equal to a semicircle, that is, to two right angles, (by the last).

Cor. 1. From whence it is plain, that all the angles which can be made from a point in any line, towards one side of the line, are equal to two right angles.

2. And that all the angles which can be made about a point, are equal to four right ones.

32. If one line AC cross another BD in the point E, then the opposite angles are equal, viz. BEA to CED, and BEC equal to AED; fig. 17. For upon the point E, as a centre, describing the circle ABCD, it is plain ABC is a semicircle, as also BCD (by the 14th); therefore the arc ABC is equal to the arc BCE; and from both taking the common arc BC, there will remain AB equal to CD, i. e. the Angle BEA equal to the angle CED (by art. 29.). After the same manner we may prove, that the angle BEC is equal to the angle AED.

33. Lines which are equally distant from one another, are called parallel lines; as AB, CD, fig. 18.

34. If a line GH cross two parallels AB, CD, (fig. 19.) then the external opposite angles are equal, viz. GEB equal to CFH and AEG equal to HFD. For since AB and CD are parallel to one another, they may be considered as one broad line, and GH crossing it; then the vertical or opposite angles GEB CFE are equal (by art. 32.), as also AEG and HFD by the same.

35. If a line GH cross two parallels AB, CD, then the alternate angles, viz. AEF and EFD, or CFE and FEB, are equal; that is, the angle AEF is equal to the angle EFD, and the angle CFE is equal to the angle FEB, for GEB is equal to AEF (by art. 32.), and CFH is equal to EFD (by the same); but GEB is equal to CFH (by the last); therefore AEF, is equal to EFD. The same way we may prove FEB equal to EFC.

36. If a line GH cross two parallel lines AB, CD, then the external angle GEB is equal to the internal opposite one EFD, or GEA equal to CFE. For the angle AEF is equal to the angle EFD (by the last); but AEF is equal to GEB (by art. 32.), therefore GEB is equal to EFD. The same way we may prove AEG equal to CFE.

37. If a line GH cross two parallel lines AB, CD, then the sum of the two internal angles, viz. BEF and DFE, or AEF and CFE, are equal to two right angles; for since the angle GEB is equal to the angle EFD (by art. 36.), to both add the angle FEB, then GEB and BEF are equal to BEF and DFE; but GEB and BEF are equal to two right angles (by art. 31.) therefore BEF and DFE are also equal to two right angles. The same way we may prove that AEF and CFE are equal to two right angles.

39. A figure is any part of space bounded by lines or a line. If the bounding lines be strait, it is called a rectilinear figure, as A, fig. 20. if they be curved, it is called a curvilinear figure, as B or C, fig. 21. and fig. 22. if they be partly curve lines and partly strait, it is called a mixed figure, as D, fig. 23.

35. The most simple rectilinear figure is that which is bounded by three right lines, and is called a triangle, as A, fig. 24.

40. Triangles are divided into different kinds, both with respect to their sides and angles: with respect to their sides they are commonly divided into three kinds, viz.

47. A triangle having all its three sides equal to one another, is called an equilateral triangle, as A, fig. 25.

42. A triangle having two of its sides equal to one another, and the third side not equal to either of them, is called an isosceles triangle, as B, fig. 26.

43. A triangle having none of its sides equal to one another, is called a scalene triangle, as C, fig. 27.

44. Triangles, with respect to their angles, are divided into three different kinds, viz.

45. A triangle having one of its angles right, is called a right-angled triangle, as A, fig. 28.

46. A triangle having one of its angles obtuse, or greater than a right angle, is called an obtuse-angled triangle, as B, fig. 29.

47. Lastly, a triangle having all its angles acute, is called an acute-angled triangle, as C, fig. 30.

48. In all right-angled triangles, the sides comprehending the right angle are called the legs, and the side opposite to the right angle is called the hypotenuse. Thus in the right-angled triangle ABC, fig. 31. (the right angle being at B) the two sides AB and BC, which comprehend the right angle ABC, are the legs of the triangle; and the side AC, which is opposite to the right angle ABC, is the hypotenuse of the right-angled triangle ABC.

49. Both obtuse and acute angled triangles are in general called oblique-angled triangles; in all which any side is called the base, and the other two the sides.

50. The perpendicular height of any triangle is a line drawn from the vertex to the base perpendicularly; thus if the triangle ABC (fig. 32.) be proposed, and BC be made its base, then A will be the vertex, viz. the angle opposite to the base; and if from A you draw the line AD perpendicular to BC, then the line AD is the height of the triangle ABC standing on BC as its base.

Hence all triangles standing between the same parallels have the same height, since all the perpendiculars are equal by the nature of parallels.

51. A figure bounded by four sides is called a quadrilateral or quadrangular figure, as ABDC, fig. 33.

52. Quadrilateral figures whose opposite sides are parallel, are called parallelograms. Thus in the quadrilateral figure ABDC, if the side AC be parallel to the side BD which is opposite to it, and AB be parallel to CD, then the figure ABDC is called a parallelogram.

53. A parallelogram having all its sides equal and angles right, is called a square, as A, fig. 34.

54. That which hath only the opposite sides equal and its angles right, is called a rectangle, as B, fig. 35.

55. That which hath equal sides but oblique angles, is called a rhombus, as C, fig. 36. and is just an inclined square.

56. That which hath only the opposite sides equal and the angles oblique, is called a rhomboides, as D, fig. 37. and may be conceived as an inclined rectangle.

57. When none of the sides are parallel to another, then the quadrilateral figure is called a trapezium.

58. Every other right lined figure, that has more sides than four, is in general called a polygon. And figures are called by particular names according to the number of their sides, viz. one of five sides is called a pentagon, of

six a hexagon, of seven a heptagon, and so on. When the sides forming the polygon are equal to one another, the figure is called a regular figure or polygon.

59. In any triangle ABC (fig. 38.) one of its legs, as BC, being produced towards D, the external angle ACD is equal to both the internal opposite ones taken together, viz. to ABC and BAC. In order to prove this, through C draw CE parallel to AB; then since CE is parallel to AB, and the lines AC and BD crosseth them, the angle ECD is equal to ABC (by art. 36.) and the angle ACE equal to CAB (by art. 35.); therefore the angles ECD and ECA are equal to the angles ABC and CAB; but the angles ECD and ECA are together equal to the angle ACD; therefore the angle ACD is equal to both the angles ABC and CAB taken together.

Cor. Hence it may be proved, that if two lines AB and CD (fig. 39.) be crossed by a third line EF, and the alternate angles AEF and EFD be equal, the lines AB and CD will be parallel; for if they are not parallel, they must meet one another on one side of the line EF (suppose at G) and so form the triangle EFG, one of whose sides GE being produced to A, the exterior angle AEF must (by this article) be equal to the sum of the two angles EFG and EGF; but, by supposition, it is equal to the angle EFG alone; therefore the angle AEF must be equal to the sum of the two angles EFG and EGF, and at the same time equal to the angle EFG alone, which is absurd; so the lines AB and CD cannot meet, and therefore must be parallel.

60. In any triangle ABC, all the three angles taken together are equal to two right angles. To prove this, you must produce BC, one of its legs, to any distance, suppose to D; then by the last proposition, the external angle, ACD, is equal to the sum of the two internal opposite ones CAB and ABC; to both add the angle ACB, then the sum of the angles ACD and ACB will be equal to the sum of the angles CAB and CBA and ACB. But the sum of the angles ACD and ACB, is equal to two right ones (by art. 32.), therefore the sum of the three angles CAB and CBA and ACB, is equal to two right angles; that is, the sum of the three angles of any triangle ACB is equal to two right angles.

Cor. 1. Hence in any triangle given, if one of its angles be known, the sum of the other two is also known: for since (by the last) the sum of all the three is equal to two right angles, or a semicircle, it is plain, that taking any one of them from a semicircle or 180 degrees, the remainder will be the sum of the other two. Thus (in the former triangle ABC) if the angle ABC be 40 degrees, by taking 40 from 180 we have 140 degrees; which is the sum of the two angles BAC, ACB: the converse of this is also plain, viz. the sum of any two angles of a triangle being given, the other angle is also known by taking that sum from 180 degrees.

2. In any right-angled triangle, the two acute angles must just make up a right one between them; consequently, any one of the oblique angles being given, we may find the other by subtracting the given one from 90 degrees, which is the sum of both.

61. If in any two triangles, ABC (fig. 40.) DEF (fig. 41.) two legs of the one, viz. AB and AC, be equal to

two legs in the other, viz. to DE and DF, each to each respectively, i. e. AB to DE, and AC to DF; and if the angles included between the equal legs be equal, viz. the angle BAC equal to the angle EDF; then the remaining leg of the one shall be equal to the remaining leg of the other, viz. BC to EF; and the angles opposite to equal legs shall be equal, viz. ABC equal to DEF (being opposite to the equal legs AC and DF), also ACB equal to DFE (which are opposite to the equal legs AB and DE). For if the triangle ABC be supposed to be lifted up and put upon the triangle DEF, and the point A on the point D; it is plain, since BA and DE are of equal length, the point E will fall upon the point B; and since the angles BAC and EDF are equal, the line AC will fall upon the line DF; and they being of equal length, the point C will fall upon the point F; and so the line BC will exactly agree with the line EF, and the triangle ABC will in all respects be exactly equal to the triangle DEF; and the angle ABC will be equal to the angle DEF, also the angle ACB will be equal to the angle DFE.

Cor. 1. After the same manner it may be proved, that if in any two triangles ABC, DEF, (see the preceding figure) two angles ABC and ACB of the one, be equal to two angles DEF and EFD of the other, each to each respectively, viz. the angle ABC to the angle DEF, and the angle ACB equal to the angle DFE, and the sides included between these angles be also equal, viz. BC equal to EF, then the remaining angles and the sides opposite to the equal angles, will also be equal each to each respectively; viz. the angle BAC equal to the angle BDF, the side AB equal to DE, and AC equal to DF: for if the triangle ABC be supposed to be lifted up and laid upon the triangle DEF, the point B being put upon the point E, and the line BC upon the line EF, since BC and EF are of equal lengths, the point C will fall upon the point F, and since the angle ACB is equal to the angle DFE, the line CA will fall upon the line FD, and by the same way of reasoning the line BA will fall upon the line ED; and therefore the point of intersection of the two lines BA and CA, viz. A, will fall upon the point of intersection of the two lines BD and FD, viz. D, and consequently BA will be equal to DE, and AC equal to DF, and the angle BAC equal to the angle EDF.

Cor. 2. It follows likewise from this article, that if any triangle ABC (fig. 42.) has two of its sides AB and AC equal to one another, the angles opposite to these sides will also be equal, viz. the angles ABC equal to the angle ACB. For suppose the line AD, bisecting the angle BAC, or dividing it into two equal angles BAD and CAD, and meeting BC in D, then the line AD will divide the whole triangle BAC into two triangles ABD and ADC; in which BA and AD two sides of the one, are equal to CA and AD two sides of the other, each to each respectively, and the included angles BAD and DAC are by supposition equal; therefore (by this article) the angle ABC must be equal to the angle ACB.

62. Any angle, as BAD (fig. 43.) at the circumference of a circle BADE, is but half the angle BCD at the centre standing on the same arch BED. To demonstrate this, draw through A and the centre C, the right

line ACE, then the angle ECD is equal to both the angles DAC and ADG (by art. 59.); but since AC and CD are equal (being two radii of the same circle) the angles subtended by them must be equal also, (by art. 62. cor. 2.) i. e. the angle CAD equal to the angle CDA; therefore the sum of them is double any one of them, i. e. DAC and ADC is double of CAD, and therefore ECD is also double of DAC: the same way it may be proved, that ECB is double of CAB; and therefore the angle BCD is double of the angle BAD, or BAD the half of BCD, which was to be proved.

Cor. 1. Hence an angle at the circumference is measured by half the arc it subtends; for the angle at the centre (standing on the same arc) is measured by the whole arc (by art. 29.); but since the angle at the centre is double that at the circumference, it is plain the angle at the circumference must be measured by only half the arc it stands upon.

Cor. 2. Hence all angles, ACB, ADB, AEB, &c. (fig. 44.) at the circumference of a circle, standing on the same chord AB, are equal to one another; for by the last corollary they are all measured by the same arc, viz. half the arc AB which each of them subtends.

Cor. 3. Hence an angle in a segment greater than a semicircle is less than a right angle: thus, if ADB be a segment greater than a semicircle, (see the last figure) than the arc AB, on which it stands, must be less than a semicircle, and the half of it less than a quadrant or a right angle; but the angle ADB in the segment is measured by the half of AB, therefore it is less than a right angle.

Cor. 4. An angle in a semicircle is a right angle. For since ABD (fig. 45.) is a semicircle, the arc AED must also be a semicircle: but the angle ABD is measured by half the arc AED, that is, by half a semicircle or quadrant; therefore the angle ABD is a right one.

Cor. 5. Hence an angle in a segment less than a semicircle, as ABD, (fig. 46.) is greater than a right angle: for since the arc ABD is less than a semicircle, the arc AED must be greater than a semicircle, and so it is half greater than a quadrant, i. e. than the measure of a right angle; therefore the angle ABD, which is measured by half the arc AED, is greater than a right angle.

63. If from the centre C of the circle ABE, (fig. 47.) there be let fall the perpendicular CD on the chord AB, then that perpendicular will bisect the chord AB in the point D. To demonstrate this, draw from the centre to the extremities of the chord the two lines CA, CB; then since the lines CA and CB are equal, the angles CAB, CBA, which they subtend must be equal also, (by art. 62. cor. 2.) but the perpendicular CD divides the triangle ACB into two right-angled triangles ACD and CBD, in which the sum of the angles ACD and CAD in the one, is equal to the sum of the angles DCB and CBD in the other, each being equal to a right angle, (by cor. 2. of art. 61.) but CAD is equal to CBD, therefore ACD is equal to BCD. So in the two triangles ACD and BCD, the two legs AC and CD in the one, are equal to the two legs BC and CD in the other, each to each respectively, and the included angles ACD and BCD are equal; therefore the remaining legs AD and BD are equal (by art. 61.) and consequently AB bisected in D.

This plate contains 53 numbered geometric diagrams and two large construction tools. The diagrams are arranged in rows and columns:

• Row 1: 1. A horizontal line segment AB with a curved line below it labeled 'a' and 'b'. 2. A square. 3. A V-shape with vertices A, B, and C. 4. A V-shape with vertices A and B. 5. A curved line segment. 6. A curved line segment. 7. A vertical line segment AB with a horizontal line segment DC below it, and a line segment from C to the right.
• Row 2: 8. A horizontal line segment CD with a vertical line segment AB above it, and a line segment from C to the right. 9. A line segment with a point B and a line extending to the left. 10. A circle with center C and points A, B, D, E on its circumference. 11. A circle with points A, B, D on its circumference. 12. A circle with center C and points A, B on its circumference. 13. A circle with center C and points A, B, D, E, F, G on its circumference. 14. A semi-circle with points A, B, C, D, E on its circumference. 15. A semi-circle with points A, B, C, D on its circumference.
• Row 3: 16. A semi-circle with points A, B, C, D on its circumference. 17. A circle with center E and points A, B, C, D on its circumference. 18. Two parallel horizontal line segments AB and CD with vertical segments AC and BD. 19. Two parallel horizontal line segments AB and CD intersected by a diagonal line GH. 20. A regular pentagon. 21. A circle. 22. A semi-circle with a curved base.
• Row 4: 23. A sector of a circle. 24. A triangle. 25. An equilateral triangle. 26. A tall, narrow triangle. 27. A triangle. 28. A right-angled triangle. 29. A right-angled triangle. 30. An equilateral triangle.
• Row 5: 31. A right-angled triangle ABC. 32. A triangle ABC with a median AD. 33. A rectangle ABCD. 34. A square. 35. A rectangle. 36. A parallelogram. 37. A parallelogram.
• Row 6: 38. Two triangles ABC and CDE sharing vertex C. 39. A complex polygonal figure with vertices A, B, C, D, E, F, G. 40. A triangle ABC. 41. A triangle DEF. 42. A triangle ABC with a median AD.
• Row 7: 43. A circle with points A, B, C, D, E on its circumference. 44. A circle with points A, B, C, D, E on its circumference. 45. A circle with points A, B, C, D, E on its circumference. 46. A circle with points A, B, C, D, E on its circumference. 47. A circle with points A, B, C, D, E on its circumference. 48. A circle with points A, B, C, D, E on its circumference. 49. A circle with points A, B, C, D, E on its circumference.
• Row 8: 50. A rectangle ABCD with a diagonal BD. 51. A rectangle ABCD with a diagonal BD and a shaded triangle BCD. 52. A complex geometric construction involving multiple lines and points. 53. Two large construction tools: 'A Line of Tangents' and 'A Line of Secants'.

64. If from the centre C of a circle ABE, there be drawn a perpendicular CD on the chord AB, and produced till it meet the circle in F, then the line CF bisects the arch AB in the point F; for (see the foregoing figure) joining the points A and F, F and B by the straight lines AF, FB, then in the triangles ADF, BDF, AD is equal to DB (by art. 63.) and DF common to both; therefore AD and DF, two legs of the triangle ADF, are equal to BD and DF, two legs of the triangle BDF, and the included angles ADF, BDF are equal, being both right; therefore (by art. 61) the remaining legs AF and FB are equal; but in the same circle equal lines are chords of equal arches, therefore the arches AF and FB are equal. So the whole arch AFB is bisected in the point F by the line CF.

Cor. 1. From art. 63. it follows, that any line bisecting a chord at right angles is a diameter; for since (by art. 63.) a line drawn from the centre perpendicular to a chord, bisects that chord at right angles; therefore, conversely, a line bisecting a chord at right angles, must pass thro' the centre, and consequently be a diameter.

Cor. 2. From the two last articles it follows, that the sine of any arc is the half of the chord of twice the arc; for (see the foregoing scheme) AD is the sine of the arc AF, by the definition of a sine, and AF is half the arc AFB, and AD half the chord AB (by art. 63.); therefore the cor. is plain.

65. In any triangle, the half of each side is the sine of the opposite angle; for if a circle be supposed to be drawn through the three angular points A, B, and D of the triangle ABD, fig. 48. then the angle DAB is measured by half the arch BKD (by cor. 1. of art. 62.) but the half of BD, viz. BE, is the sine of half the arch BKD, viz. the sine of BK (by cor. 2. of the last) which is the measure of the angle BAD; therefore the half of BD is the sine of the angle BAD; the same way it may be proved, that the half of AD is the sine of the angle ABD, and the half of AB is the sine of the angle ADB.

66. The sine, tangent, &c. of any arch is called also the sine, tangent, &c. of the angle whose measure the arc is: thus because the arc GD (fig. 49.) is the measure of the angle GCD; and since GH is the sine, DE the tangent, HD the versed sine, CE the secant, also GK the co-sine, BF the co-tangent, and CF the co-secant, &c. of the arch GD; then GH is called the sine, DE the tangent, &c. of the angle GCD, whose measure is the arch GD.

67. If two equal and parallel lines, AB and CD (fig. 50.) be joined by two others, AC and BD; then these shall also be equal and parallel. To demonstrate this, join the two opposite angles A and D with the line AD; then it is plain this line AD divides the quadrilateral, ACDB, into two triangles, viz. ABD, ACD, in which AB a leg of the one, is equal to DC a leg of the other, by supposition, and AD is common to both triangles; and since AB is parallel to CD, the angle BAD will be equal to the angle ADC, (by art. 36.) therefore in the two triangles BA and AD; and the angle BAD is equal to CD and DA; and the angle ADC, that is, two legs and the included angle in the one, is equal to two legs and the included angle in the other; therefore (by art. 61) BD is equal to AC, and since the angle DAC

is equal to the angle ADB, therefore the lines BD AC are parallel (by cor. art. 59.)

Cor. 1. Hence it is plain, that the quadrilateral ABDC is a parallelogram, since the opposite sides are parallel.

Cor. 2. In any parallelogram the line joining the opposite angles (called the diagonal) as AD, divides the figure into two equal parts, since it has been proved that the triangles ABD ACD are equal to one another.

Cor. 3. It follows also, that a triangle ACD on the same base CD, and between the same parallels with a parallelogram ABDC, is the half of that parallelogram.

Cor. 4. Hence it is plain, that the opposite sides of a parallelogram are equal; for it has been proved, that ABDC being a parallelogram, AB will be equal to CD, and AC equal to BD.

68. All parallelograms on the same or equal bases, and between the same parallels, are equal to one another; that is, if BD and GH (fig. 51.) be equal, and the lines BH and AF be parallel, then the parallelograms ABDC, BDFE, and EFHG, are equal to one another. For AC is equal to EF, each being equal to BD, (by cor. 4. of 67.) To both add CE, then AE will be equal CF. So in the two triangles ABE CDF, AB a leg of the one, is equal to CD a leg in the other; and AE is equal to CF, and the angle BAE is equal to the angle DCF (by art. 36.); therefore the two triangles ABE CDF are equal (by art. 61.); and taking the triangle CKE from both, the figure ABKC will be equal to the figure KDFE; to both which add the little triangle KBD, then the parallelogram ABDC will be equal to the parallelogram BDFE. The same way it may be proved, that the parallelogram EFHG is equal to the parallelogram EFDB; so the three parallelograms ABDC, BDFE, and EFHG will be equal to one another.

Cor. Hence it is plain, that triangles on the same base, and between the same parallels, are equal; since they are the half of the parallelograms on the same base and between the same parallels. (by cor. 3. of last art.)

69. In any right-angled triangle, ABC, (fig. 52.) the square of the hypotenuse BC, viz. BCMH, is equal to the sum of the squares made on the two sides AB and AC, viz. to ABDE and ACGF. To demonstrate this, through the point A draw AKL perpendicular to the hypotenuse BC, join AH, AM, DC, and BG; then it is plain that DB is equal to BA (by art. 53.), also BH is equal to BC (by the same); so in the two triangles DBC ABH, the two legs DB and BC in the one are equal to the two legs AB and BH in the other; and the included angles DBC and ABH are also equal; (for DBA is equal to CBH, being both right; to each add ABC, then it is plain that DBC is equal to ABH) therefore the triangles DBC ABH are equal (by art. 61.) but the triangle DBC is half of the square ABDE (by cor. 3. of 67th) and the triangle ABH is half the parallelogram BKLH (by the same), therefore half the square ABDE is equal to half the parallelogram BKLH. Consequently the square ABDE is equal to the parallelogram BKLH. The same way it may be proved, that the square ACGF is equal to the parallelogram KCML. So the sum of the squares ABDE and ACGF is equal to the sum of the parallelograms BKLH and KCML, but the sum of these parallelograms is equal to the square BCMH, therefore the

sum of the squares on AB and AC is equal to the square on BC.

Cor. 1. Hence in a right-angled triangle, the hypothenuse and one of the legs being given, we may easily find the other, by taking the square of the given leg from the square of the hypothenuse, and the square root of the remainder will be the leg required.

Cor. 2. Hence, the legs in a right-angled triangle being given, we may find the hypothenuse, by taking the sum of the squares of the given legs, and extracting the square root of that sum.

70. If upon the line AB (fig. 53.) there be drawn a semicircle ADB, whose centre is C, and on the point C there be raised a perpendicular to the line AB, viz. CD; then it is plain the arc DB is a quadrant, or contains 90 degrees; suppose the arc DB to be divided into 9 equal arcs, each of which will contain 10 degrees, then on the point B raising BE perpendicular to the line AB, it will be a tangent to the circle in the point B, and if to every one of the divisions of the quadrant, viz. B 10, B 20, B 30, B 40, &c. you draw the sine, tangent, &c. (as in the scheme) we shall have the sine, tangent, &c. to every ten degrees in the quadrant: and the same way we may have the sine, tangent, &c. to every single degree in the quadrant, by dividing it into 90 equal parts beginning from B, and drawing the sine, tangent, &c. to all the arcs beginning at the same point B. By this method they draw the lines of sines, tangents, &c. of a certain circle on the scale; for after drawing them on the circle, they take the length of them, and set them off in the lines drawn for that purpose. The same way, by supposing the radius of any number of equal parts, (suppose 1000, or 10,000, &c.) it is plain the sine, tangent, &c. of every arc must consist of some number of these equal parts; and by computing them in parts of the radius, we have tables of sines, tangents, &c. to every arc in the quadrant, called natural sines, tangents, &c. and the logarithms of these give us tables of logarithmic sines, tangents, &c. See LOGARITHMS.

71. In any triangle, ABC. (Plate XCIV. fig. 1.) if one of its sides, as AC, be bisected in E, (and consequently AC double of AE) and through E be drawn ED, parallel to BC, and meeting AB in D, then BC will be double of ED, and AB double of AD, through D draw DF, parallel to AC, meeting BC in F: for since, by construction, DF is parallel to AC, and DE parallel to BC; therefore, (by art. 36.) the angle BFD will be equal to the angle BCA, (and by the same article) the angle BCA will be equal to the angle DEA, consequently the angle BFD will be equal to the angle DEA; also, (by art. 36.) the angle BDF will be equal to the angle DAE; and since DF is parallel to EC, and DB parallel to FC, the quadrilateral DFCE will be a parallelogram; and therefore, (by art. 59. cor. 4.) DF will be equal to EC, which, by construction, is equal to AE; so in the two triangles BDF and DAE, the two angles BFD and BDF in the one, are equal to the two angles DBA and DAE in the other, each to each respectively; and the included side DF, is equal to the included side AE; therefore, (by art. 61. cor. 1.) AD will be equal to DB, and consequently AB double of AD;

also (by the same) DE will be equal to BF; but DE is also (by art. 67. cor. 4.) equal to FC; therefore BF and EC together, or BC, will be double of DE.

After the same manner it may be proved, that if in the triangle AKG, (fig. 2.) AE be taken equal to a third part of AK, and through E be drawn ED, parallel to KG, and meeting AG in D; then ED will be equal to a third part of GK, and AD equal to a third part of AG.

Likewise if in any triangle ABC, (fig. 3.) upon the side AB, be taken AE, equal to one fourth, one fifth, one sixth, &c. of AB, and through E be drawn ED parallel to BC and meeting AC in D; then DE will be one fourth, one fifth, one sixth, &c. of BC, and AD the like part of AC; and, in general, if in any triangle ABC, there be assumed a point E on one of its sides AB, and through that point be drawn a line ED, parallel to one of its sides BC, and meeting the other side AC in D; then whatever part AE is of AB, the same part will ED be of BC, and AD of AC.

Cor. Hence it follows, that if in any triangle ABC, there be drawn ED, parallel to one of its sides BC, and meeting the other two in the points E and D, then $AE\; :\; AB\; ::\; ED\; :\; BC\; ::\; AD\; :\; AC$; that is, AE is to AB, as ED is to BC, and that as AD to AC.

72. If any two triangles ABC, fig. 4. abc, fig. 5. are similar, or have all the angles of the one equal to all the angles of the other, each to each respectively; that is, the angle CAB equal to the angle cab, and the angle ABC equal to the angle abc, and the angle ACB equal to the angle acb; then the legs opposite to the equal angles are proportioned, viz. $AB\; :\; ab\; ::\; AC\; :\; ac\; ::$ and $AB\; :\; ab\; ::\; BC\; :\; bc\; ::$ and $AC\; :\; ac\; ::\; BC\; :\; bc$. On AB of the largest triangle set off AE equal to ab, and through E draw ED parallel to BC, meeting AC in D; then since DE and BC are parallel, and AB crossing them, the angle AED will (by art. 36.) be equal to the angle ABC, which (by supposition) is equal to the angle abc, also the angle DAE is (by supposition) equal to the angle cab; so in the two triangles AED, abc, the two angles DAE AED of the one, are equal to two angles cab abc of the other, each to each respectively, and the included side AE is (by construction) equal to the included side ab; therefore, (by art. 61. cor. 1.) AD is equal to ac, and DE equal to cb; but since, in the triangle ABC, there is drawn DE parallel to BC one of its sides, and meeting the two other sides in the points D and E, therefore (by cor. art. 71.) $AB\; :\; AE\; ::\; AC\; :\; AD$, and $AB\; :\; AE\; ::\; BC\; :\; DE$, and $AC\; :\; AD\; ::\; BC\; :\; DE$; and in the three last proportions, instead of the lines AE, DE, and AD, putting in their equals ab, bc, and ac, we shall have $AB\; :\; ab\; ::\; AC\; :\; ac$, and $AB\; :\; ab\; ::\; BC\; :\; bc$, and lastly, $AC\; :\; ac\; ::\; BC\; :\; bc$.

73. The chord, sine, tangent, &c. of any arc in one circle, is to the chord, sine, tangent, &c. of the same arc in another, as the radius of the one is to the radius of the other, fig. 6, 6. Let ABD abd be two circles, BD bd two arcs of these circles, equal to one another, or consisting of the same number of degrees; FD fd the tangents, BD bd the chords, BE be the sines,

Ec. of these two arcs BD bd, and CD cd the radii of the circles; then say, $CD\; :\; cd\; ::\; FD\; :\; fd$, and $CD\; :\; cd\; ::\; BD\; :\; bd$, and $CD\; :\; cd\; ::\; BE\; :\; be$, &c. For since the arcs BD bd are equal, the angles BCD bcd will be equal; and FD fd, being tangents to the points D and d, the angles CDF cdf will be equal, being each a right angle (art. 22.) so in the two triangles CDF cdf, the two angles FCD CDF of the one, being equal to the two angles fcd cdf of the other, each to each, the remaining angle CFD, will be equal to the remaining angle cfd, (by art. 60.); therefore the triangles CFD cfd are similar, and consequently (by art. 73.) $CD\; :\; cd\; ::\; FD\; :\; fd$. In the same manner it may be demonstrated, that $CD\; :\; cd\; ::\; BD\; :\; bd$, and $CD\; :\; cd\; ::\; BE\; :\; be$, &c.

74. Let ABD (fig. 7.) be a quadrant of a circle described by the radius CD BD any arc of it, and BA its complement, BG or CF the sine, CG or BF the cosine, DE the tangent, and CE the secant of that arc BD. Then since the triangles CDE CGB are similar or equiangular, it will be (by art. 72.) $DE\; :\; EC\; ::\; GB\; :\; BC$, i. e. the tangent of any arc, is to the secant of the same, as the sine of it is to the radius. Also since $DE\; :\; EC\; ::\; GB\; :\; BC$; therefore, by inverting that proportion, we have $EC\; :\; DE\; ::\; BC\; :\; GB$, i. e. the secant is to the tangent, as the radius is to the sine of any arc.

Again, since the triangles CDE CGB are similar, therefore (by art. 72.) it will be $CD\; :\; CE\; ::\; CG\; :\; CB$, i. e. as the radius is to the secant of any arc, so is the cosine of that arc to the radius. And by inverting the proportion we have this, viz. As the secant of any arc is to the radius, so is the radius to the cosine of that arc.

75. In all circles the chord of 60 is always equal in length to the radius. Thus in the circle AEBD, (fig. 8.) if the arc AEB be an arc of 60 degrees, then drawing the chord AB, I say AB shall be equal to the radius CB or AC; for in the triangle ACB, the angle ACB is 60 degrees, being measured by the arc AEB; therefore the sum of the other two angles is 120 degrees, (by cor. 1. of 60.); but since AC and CB are equal to the two angles CAB, CBA will also be equal; consequently each of them half their sum 120, viz. 60 degrees; therefore all the three angles are equal to one another, consequently all the legs, therefore AB is equal to CB.

Cor. Hence the radius from which the lines on any scale are formed, is the chord of 60 on the line of chords.

Geometrical Problems.

PROB. 1. From a point C (fig. 9.) in a given line AB to raise a perpendicular to that line.

Rule. From the point C take the equal distances CB, CA on each side of it. Then stretch the compasses to any distance greater than CB or CA, and with one foot of them in B, sweep the arc EF with the other; again, with the same opening, and one foot in A, sweep the arc GH with the other, and these two arcs will intersect one another in the point D; then join the given points C and D with the line CD, and that shall be the perpendicular required.

2. To divide a given right line AB (fig. 10.) into two equal parts; that is, to bisect it.

Rule. Take any distance with your compasses that you are sure is greater than half the given line; then setting one foot of them in B, with the other sweep the arc DFC; and with the same distance, and one foot in A, with the other sweep the arc CED; these two arcs will intersect one another in the points C D, which joined by the right line DC will bisect AB in G.

3. From a given point D, (fig. 11.) to let fall a perpendicular on a given line AB.

Rule. Set one foot of the compasses in the point D, and extend the other to any distance greater than the least distance between the given point and the line, and with that extent sweep the arc AEB, cutting the line in the two points A and B, then (by the last prob.) bisect the line AB in the point C, lastly join C and D, and that line CD is the perpendicular required.

4. (Fig. 12.) Upon the end B of a given right line BA, to raise a perpendicular.

Rule. Take any extent in your compasses, and with one foot in B fix the other in any point C without the given line; then with one point of the compasses in C, describe with the other the circle EBD, and thro' E and C draw the diameter ECD meeting the circle in D; join D and B, and the right line DB is that required; for EBD is a right angle (by cor. 4. of 63.).

5. (Fig. 13.) To draw one line parallel to another given line AB, that shall be distant from one another by any given distance D.

Rule. Extend your compasses to the given distance D; then setting one foot of them in any point of the given line (suppose A), with the other sweep the arc FCG; again, at the same extent, and one foot in any other point of the given line B, sweep the arc HDK, and draw the line CD touching them, and that will be parallel to the given line AB, and distant from it by the line D as was required.

6. (Fig. 14.) To divide a given line AB into any number of equal parts, suppose 7.

Rule. From the point A draw any line AD, making an angle with the line AB, then through the point B, draw a line BC parallel to AD; and from A, with any small opening of the compasses, set off a number of equal parts (on the line AD,) less by one than the proposed number (here 6.); then from B set off the same number of the same parts (on the line BC); lastly, join 6 and 1, 2 and 5, 3 and 4, 4 and 3, 5 and 2, 6 and 1, and these lines will cut the given line as required.

7. (Fig. 15.) To quarter a given circle, or to divide it into four equal parts.

Rule. Thro' the centre C of the given circle, draw a diameter AB, then upon the point C raise a perpendicular DCE to the line AB; and these two diameters AB and DE shall quarter the circle.

8. (Fig. 16.) Thro' three given points A, B, and D, to draw a circle. (Note. The three points must not lie in the same straight line.)

Rule. Join A and B, also B and D, with the straight lines AB BD; then (by prob. 2.) bisect AB with the line

line EC, also BD with the line FC, which two lines will cut one another in some point C; that is the centre of the circle required: then fixing one point of your compasses in D, and stretching the other to A, describe the circle ABDG, which will pass thro' the three points given. The reason of this is plain from cor. 1. of art. 64.

9. (Fig. 17.) From the point A of the given line AB, to draw another line (suppose AC) that shall make with AB an angle of any number of degrees, suppose 45.

Rule. Let the given line AB be produced, then take off your scale the length of the chord of 60 degrees, which is equal to the radius of the circle the scale was made for (by art. 75.); and setting one foot in A, with the other sweep of the arc BC; then with your compasses take from your scale the chord of 45 degrees, and set off that distance from B to C. Lastly, join A and C, and the line AC is that required. For the angle CAB, which is measured by the arc BC, is an angle of 45 degrees, as was required.

10. An angle BAC (fig. 18.) being given, to find how many degrees it contains.

Rule. With your compasses take the length of your chord of 60 from your scale. Then, setting one foot of them in A, with the other sweep the arc BC, which is the arc comprehended between the two legs AB, AC produced if needful. Lastly, take with your compasses the distance BC, and applying it to your line of chords on the scale, you will find how many degrees the arc BC contains, and consequently the degrees of the angle BAC which was required.

11. Three lines $x$, $y$, and $z$ being given, (fig. 19. 19.) to form a triangle of them; but any two of these lines taken together must always be greater than the third.

Rule. Make any one of them, as $x$, the base; then with your compasses take another of them, as $z$, and setting one foot in one end of the line $x$, as B, with the other sweep the arc DE; and taking with your compasses the length of the other $y$, set one foot of them in A, the other end of the line $x$, and with the other sweep the arc FG, which will cut the other in C; lastly, join CA and CB, and the triangle CAB is that required.

12. To make a triangle, having one of its legs of any number of equal parts (suppose 160), and one of the angles at that leg 50 degrees, and the other 44 degrees.

Rule. Draw an indefinite line ED, (fig. 20.) then take off the line of equal parts with your compasses, 160 of them, and set them on the indefinite line, as BC; then (by prob. 9.) draw BA, making the angle ABC of 50 degrees, and (by the same) draw from C the line AC, ma-

king the angle ACB of 44 degrees; which two lines will meet one another in A, and the triangle ABC is that required. See TRIGONOMETRY.

13. Upon a given line AB (fig. 21.) to make a square.

Rule. Upon the extremity A of the given line AB, raise a perpendicular AC (by prob. 4.); then take AC equal to AB, and with that extent, setting one foot of the compasses in C, sweep with the other foot the arc GH; then with the same extent, and one foot in B, with the other sweep the arc EF, which will meet the former in some point D; lastly, join C and D, D and B, and the figure ABDC will be the square required.

14. On a given line AB (fig. 22.) to draw a rhomb that shall have one of its angles equal to any number of degrees, suppose 60 degrees.

Rule. From the point A of the given line AB, draw the line AC, making the angle CAB of 60 degrees, (by prob. 9.) then take AC equal to AB, and with that extent, fixing one foot of the compasses in B, with the other describe the arc GH; and at the same extent, fixing one foot of the compasses in C, with the other describe the arc EF cutting the former in D; lastly, join CD and DB, and the figure ACDB is that required.

15. Given two lines $x$ and $z$, of these two to make a rectangle.

Rule. Draw a line, as AB, (fig. 23. 23.) equal in length to one of the given lines $x$; and on the extremity A of that line, raise a perpendicular AC, on which take AC equal to the other line $z$; then take with your compasses the length of the line AB, and at that extent, fixing one foot of them in C, with the other sweep the arc EF; and also taking with your compasses the extent of the line AC, fix one foot of them in B, and with the other sweep the arc GH, which will meet the former in D; lastly, join CD and BD, and the figure ABDC will be that required.

16. Two lines $x$ and $z$ being given, of these to form a rhomboides that shall have one of its angles any number of degrees, suppose 50.

Rule. Draw a line AB (fig. 24. 24.) equal in length to one of the lines, as $x$; then draw the line AC, making with the former the angle BAC equal to the proposed, suppose 50 degrees, and on that line take AC equal to the given line $z$, then with your compasses take the length of AB, and fixing one foot in C, sweep the arc EF; also, taking the length of AC, and setting one foot in B, with the other sweep the arc GH, which will cut the former in D; then join CD and DB, so the figure ACDB will be that required.

P A R T II.

THE APPLICATION OF THE FOREGOING PRINCIPLES TO THE MENSURATION OF SURFACES, SOLIDS, &c.

Of the Mensuration of Lines and Angles.

A Line, or length, to be measured, whether it be distance, height, or depth, is measured by a line less than it. With us the least measure of length is an inch:

not that we measure no line less than it, but because we do not use the name of any measure below that of an inch; expressing lesser measures by the fractions of an inch: and in this treatise we use decimal fractions as the easiest. Twelve inches make a foot; three feet and 20

Fig. 1.

Plate XCIV.

The plate contains 24 numbered diagrams illustrating geometric constructions and properties:

• Fig. 1: A right-angled triangle ABC with altitude AD and perpendicular DF.
• Fig. 2: A triangle ABC with a line DE parallel to the base BC.
• Fig. 3: A triangle ABC with a line DE parallel to the base BC.
• Fig. 4: A triangle ABC with a line DE parallel to the base BC.
• Fig. 5: A triangle ABC with a line DE parallel to the base BC.
• Fig. 6: A circle with a chord CD and a perpendicular BE.
• Fig. 7: A circle with a chord CD and a perpendicular BE.
• Fig. 8: A circle with a chord AB and a perpendicular CD.
• Fig. 9: A horizontal line with points A, C, B and a vertical line with points G, D, P, E, H.
• Fig. 10: A vertical line with points C, G, B, D and a horizontal line with points A, P, G, B.
• Fig. 11: A vertical line with points D, C, E and a horizontal line with points A, C, B.
• Fig. 12: A right-angled triangle ABC with a line DE parallel to the base AB.
• Fig. 13: Two parallel horizontal lines with points A, B, C, D, E, F, G, H, K.
• Fig. 14: A parallelogram with diagonal lines and numbered points 1 through 6.
• Fig. 15: A circle with a vertical diameter CD and a horizontal diameter AB.
• Fig. 16: A circle with a chord CD and a perpendicular BE.
• Fig. 17: A triangle ABC with an angle of 60 degrees at vertex A.
• Fig. 18: A triangle ABC with an angle of 25 degrees at vertex A.
• Fig. 19: A triangle ABC with angles labeled x, y, z.
• Fig. 20: A geometric construction with points A, B, C, D, E, F, G and angles of 30 degrees and 60 degrees.
• Fig. 21: A square ABCD with a line segment GH and points P, Q, R.
• Fig. 22: A parallelogram ABCD with an angle of 60 degrees at vertex A.
• Fig. 23: A rectangle ABCD with points E, F, G, H and a line segment GH.
• Fig. 24: A parallelogram ABCD with an angle of 50 degrees at vertex A.

A.D. 16

A row of six geometric diagrams. From left to right: 1. A circle with a vertical diameter and a horizontal chord. 2. A triangle with a vertical line from the top vertex to the base. 3. A triangle with a vertical line from the top vertex to the base. 4. A triangle with a vertical line from the top vertex to the base. 5. A triangle with a vertical line from the top vertex to the base. 6. A triangle with a vertical line from the top vertex to the base.

A row of five geometric diagrams. From left to right: 1. A circle with a vertical diameter and a horizontal chord. 2. A circle with a vertical diameter and a horizontal chord. 3. A circle with a vertical diameter and a horizontal chord. 4. A circle with a vertical diameter and a horizontal chord. 5. A circle with a vertical diameter and a horizontal chord.

A row of five geometric diagrams. From left to right: 1. A circle with a vertical diameter and a horizontal chord. 2. A circle with a vertical diameter and a horizontal chord. 3. A circle with a vertical diameter and a horizontal chord. 4. A circle with a vertical diameter and a horizontal chord. 5. A circle with a vertical diameter and a horizontal chord.

A row of five geometric diagrams. From left to right: 1. A circle with a vertical diameter and a horizontal chord. 2. A circle with a vertical diameter and a horizontal chord. 3. A circle with a vertical diameter and a horizontal chord. 4. A circle with a vertical diameter and a horizontal chord. 5. A circle with a vertical diameter and a horizontal chord.

A row of five geometric diagrams. From left to right: 1. A circle with a vertical diameter and a horizontal chord. 2. A circle with a vertical diameter and a horizontal chord. 3. A circle with a vertical diameter and a horizontal chord. 4. A circle with a vertical diameter and a horizontal chord. 5. A circle with a vertical diameter and a horizontal chord.

inch make the Scots ell; six ells make a fall; forty falls make a furlong; eight furlongs make a mile: so that the Scots miles is 1184 paces, accounting every pace to be five feet. These things are according to the statutes of Scotland; notwithstanding which, the glaziers use a foot of only eight inches; and other artists for the most part use an English foot, on account of the several scales marked on the English foot-measure for their use. But the English foot is somewhat less than the Scots; so that 185 of these make 186 of those.

Lines, to the extremities and any intermediate point of which you have easy access, are measured by applying to them the common measure a number of times. But lines, to which you cannot have such access, are measured by methods taken from geometry; the chief whereof we shall here endeavour to explain. The first is by the help of the geometrical square.

"As for the English measures, the yard is three feet, or thirty-six inches. A pole is sixteen feet and a half, or five yards and a half. The chain, commonly called Gunter's chain, is four poles, or twenty-two yards, that is, sixty six feet. An English statute-mile is four score chains, or 1760 yards, that is, 5280 feet.

"The chain (which is now much in use, because it is very convenient for surveying) is divided into a hundred links, each of which is $\backslash frac\{7\}\{100\}$ of an inch: whence it is easy to reduce any number of those links to feet, or any number of feet to links.

"A chain that may have the same advantages in surveying in Scotland, as Gunter's chain has in England, ought to be in length seventy-four feet, or twenty-four Scots ells, if no regard is had to the difference of the Scots and English foot above mentioned. But, if regard is had to that difference, the Scots chain ought to consist of $74\backslash frac\{1\}\{2\}$ English feet, or 74 feet 4 inches and $\backslash frac\{1\}\{2\}$ of an inch. This chain being divided into a hundred links, each of those links is 8 inches and $\backslash frac\{1\}\{100\}$ of an inch. In the following table, the most noted measures are expressed in English inches and decimals of an inch."

PROB. To describe the structure of the geometrical square.—The geometrical square is made of any solid matter, as brass or wood, or of any four plain rulers joined together at right angles, (as in Plate XCV. fig. 1.) where A is the centre, from which hangs a thread with a small weight at the end, so as to be directed always to the centre. Each of the sides BE and DE is divided into an hundred equal parts, or (if the sides be long enough to admit of it) into a thousand parts; C and F are two sights, fixed on the side AD. There is moreover an index GH, which, when there is occasion, is joined to the centre A, in such manner as that it can move round, and remain in any given situation. On this index are two sights perpendicular to the right line going from the centre of the instrument: these are K and L. The side DE of the instrument is called the upright side; E the reclining side.

FIG. 2. To measure an accessible height, AB, by the help of a geometrical square, its distance being known.—Let BR be an horizontal plane, on which there stands perpendicularly any line AB: let BD, the given distance of the observator from the height, be 96 feet; let the height of the observator's eye be supposed 6 feet; and let the instrument, held by a steady hand, or rather leaning on a support, be directed towards the summit A, so that one eye (the other being shut) may see it clearly through the sights; the perpendicular or plumb-line meanwhile hanging free, and touching the surface of the instrument

strument: let now the perpendicular be supposed to cut off on the right side $KN$ 80 equal parts. It is clear that $LKN$, $ACK$, are similar triangles; for the angles $LKN$, $ACK$ are right angles, and therefore equal; moreover $LN$ and $AC$ are parallel, as being both perpendicular to the horizon; consequently, (by art. 60. cor. 1. part 1.) the angles $KLN$, $KAC$, are equal; wherefore, (by art. 60. cor. 2. of part 1.) the angles $LNK$, and $AKC$, are likewise equal: so that in the triangles $NKL$, $KAC$, (by art. 72. of part 1.) as $NK\; :\; KL\; ::\; KC\; (i.\; e.\; BD)\; :\; CA$; that is, as 80 to 100, so is 96 feet to $CA$. Therefore, by the rule of three, $CA$ will be found to be 120 feet; and $CB$, which is 6 feet, being added, the whole height is 126 feet.

But if the distance of the observator from the height, as $BE$, be such, that, when the instrument is directed as formerly toward the summit $A$, the perpendicular falls on the angle $P$, opposite to $H$, the centre of the instrument, and $BE$ or $CG$ be given of 120 feet; $CA$ will also be 120 feet. For in the triangles $HGP$, $ACG$, equiangular, as in the preceding case, as $PG\; :\; GH\; ::\; GC\; :\; CA$. But $PG$ is equal to $GH$; therefore $GC$ is likewise equal to $CA$: that is, $CA$ will be 120 feet, and the whole height 126 feet as before.

Let the distance $BF$ be 300 feet, and the perpendicular or plumb line cut off 40 equal parts from the reclining side: Now, in this case, the angles $QAC$, $QZI$, are equal, and the angles $QZI$, $ZIS$, are equal; therefore the angle $ZIS$ is equal to the angle $QAC$. But the angles $ZSI$, $QCA$ are equal, being right angles; therefore, in the equiangular triangles $ACQ$, $SZI$, it will be, as $ZS\; :\; SI\; ::\; CQ\; :\; CA$; that is, as 100 to 40, so is 300 to $CA$. Wherefore, by the rule of three, $CA$ will be found to be of 102 feet. And, by adding the height of the observator, the whole $BA$ will be 126 feet. Note, that the height is greater than the distance, when the perpendicular cuts the right side, and less if it cut the reclined side; and that the height and distance are equal, if the perpendicular fall on the opposite angle.

SCHOLIUM.

If the height of a tower, to be measured as above, end in a point, (as in fig. 3.) the distance of the observator opposite to it, is not $CD$, but is to be accounted from the perpendicular to the point $A$; that is, to $CD$ must be added the half of the thickness of the tower, viz. $BD$: which must likewise be understood in the following propositions, when the case is similar.

PROPOSITION III.

FIG. 4. From the height of a tower $AB$ given, to find a distance on the horizontal plane $BC$, by the geometrical square.—Let the instrument be so placed, as that the mark $C$ in the opposite plane may be seen through the sights; and let it be observed how many parts are cut off by the perpendicular. Now, by what hath been already demonstrated, the triangles $AEF$, $ABC$, are similar; therefore, it will be as $EF$ to $AE$, so $AB$ (composed of the height of the tower $BG$, and of the height of the centre of the instrument $A$, above the tower $BG$) to the distance $BC$. Wherefore, if, by the rule of three, you

say, as $EF$ to $AE$, so is $AB$ to $BC$, it will be the distance sought.

PROPOSITION IV.

FIG. 5. To measure any distance at land or sea, by the geometrical square.—In this operation, the index is to be applied to the instrument, as was shown in the description; and, by the help of a support, the instrument is to be placed horizontally at the point $A$; then let it be turned till the remote point $F$, whose distance is to be measured, be seen through the fixed sights; and bringing the index to be parallel with the other side of the instrument, observe by the sights upon it any accessible mark $B$, at a sensible distance: then carrying the instrument to the point $B$, let the immovable sights be directed to the first station $A$, and the sights of the index to the point $F$. If the index cut the right side of the square, as in $K$, in the two triangles $BRK$, and $BAF$, which are equiangular, it will be as $BR$ to $RK$, so $BA$ (the distance of the stations to be measured with a chain) to $AF$; and the distance $AF$ sought will be found by the rule of three. But if the index cut the reclined side of the square in any point $L$, where the distance of a more remote point is sought; in the triangles $BLS$, $BAG$, the side $LS$ shall be to $SB$, as $BA$ to $AG$, the distance sought; which accordingly will be found by the rule of three.

PROPOSITION V.

FIG. 6. To measure an accessible height by means of a plain mirror.—Let $AB$ be the height to be measured; let the mirror be placed at $C$, in the horizontal plane $BD$, at a known distance $BC$; let the observer go back to $D$, till he see the image of the summit in the mirror, at a certain point of it, which he must diligently mark; and let $DE$ be the height of the observator's eye. The triangles $ABC$ and $EDC$ are equiangular; for the angles at $D$ and $B$ are right angles; and $ACB$, $EDC$, are equal, being the angles of incidence and reflexion of the ray $AC$, as is demonstrated in optics; wherefore the remaining angles at $A$ and $E$ are also equal: therefore it will be, as $CD$ to $DE$, so $CB$ to $BA$; that is, as the distance of the observator from the point of the mirror in the right line betwixt the observator and the height, is to the height of the observator's eye, so is the distance of the tower from that point of the mirror, to the height of the tower sought; which therefore will be found by the rule of three.

Note 1. The observation will be more exact, if, at the point $D$, a staff be placed in the ground perpendicularly, over the top of which the observator may see a point of the glass exactly in a line betwixt him and the tower.

Note 2. In place of a mirror may be used the surface of water contained in a vessel, which naturally becomes parallel to the horizon.

PROPOSITION VI.

FIG. 7. To measure an accessible height $AB$ by means of two staves.—Let there be placed perpendicularly in the ground a longer staff $DE$, likewise a shorter one $FG$, so as the observator may see $A$, the top of the height to be measured, over the ends $D$ and $F$ of the two staves; let $EH$ and $DC$, parallel to the horizon, meet $DE$ and $AB$ in $H$ and $C$; then the triangles $FHD$, $DCA$, shall be equiangular;

equiangular; for the angles at C and H are right ones; likewise the angle A is equal to the angle FDH; wherefore the remaining angles DFH, and ADC, are also equal: Wherefore, as FH, the distance of the staves, to HD, the excess of the longer staff above the shorter; so is DC, the distance of the longer staff from the tower, to CA, the excess of the height of the tower above the longer staff. And thence CA will be found by the rule of three.

To which if the length DE be added, you will have the whole height of the tower BA.

S C H O L I U M.

FIG. 8. Many other methods may be occasionally contrived for measuring an accessible height. For example, from the given length of the shadow BD, to find out the height AB, thus: Let there be erected a staff CE perpendicularly, producing the shadow EF: The triangles ABD, CEF, are equiangular; for the angles at B and E are right; and the angles ADB and CFE are equal, each being equal to the angle of the sun's elevation above the horizon: Therefore, as EF, the shadow of the staff, to EC, the staff itself; so BD, the shadow of the tower, to BA, the height of the tower. Though the plane on which the shadow of the tower falls be not parallel to the horizon, if the staff be erected in the same plane, the rule will be the same.

P R O P O S I T I O N VII.

To measure an inaccessible height by means of two staves.—Hitherto we have supposed the height to be accessible, or that we can come at the lower end of it; now if, because of some impediment, we cannot get to a tower, or if the point whose height is to be found out be the summit of a hill, so that the perpendicular be hid within the hill; if, for want of better instruments, such an inaccessible height is to be measured by means of two staves, let the first observation be made with the staves DE and FG, (as in prop. 6.); then the observator is to go off in a direct line from the height and first station, till he come to the second station; where (fig. 11.) he is to place the longer staff perpendicularly at RN, and the shorter staff at KO, so that the summit A may be seen along their tops; that is, so that the points KNA may be in the same right line. Through the point N, let there be drawn the right line NP parallel to FA: Wherefore in the triangles KNP, KAF, the angles KNP, KAF are equal, also the angle AKF is common to both; consequently the remaining angle KPN is equal to the remaining angle KFA. And therefore, PN : FA :: KP : KF. But the triangles PNL, FAS are similar; therefore, PN : FA :: NL : SA. Therefore, (by the 11. 5. Eucl.) KP : KF :: NL : SA. Thence, alternately, it will be, as KP (the excess of the greater distance of the short staff from the long one above its lesser distance from it) to NL, the excess of the longer staff above the shorter; so KF, the distance of the two stations of the shorter staff to SA the excess of the height sought above the height of the shorter staff. Wherefore SA will be found by the rule of three. To which let the height of the shorter staff be added, and the sum will give the whole inaccessible height BA.

Note 1. In the same manner may an inaccessible

height be found by a geometrical square, or by a plain speculum. But we shall leave the rules to be found out by the student, for his own exercise.

Note 2. That by the height of the staff we understand its height above the ground in which it is fixed.

Note 3. Hence depends the method of using other instruments invented by geometers; for example, of the geometrical cross: And if all things be justly weighed, a like rule will serve for it as here. But we incline to touch only upon what is most material.

P R O P O S I T I O N VIII.

FIG. 9. To measure the distance AB, to one of whose extremities we have access, by the help of four staves.—Let there be a staff fixed at the point A; then going back at some sensible distance in the same right line, let another be fixed in C, so as that both the points A and B be covered and hid by the staff C: likewise going off in a perpendicular from the right line CB, at the point A, (the method of doing which shall be shown in the following scholium), let there be placed another staff at H; and in the right line CKG (perpendicular to the same CB, at the point B), and at the point of it K, such that the points K, H, and B may be in the same right line, let there be fixed a fourth staff. Let there be drawn, or let there be supposed to be drawn, a right line GH parallel to CA. The triangles KGH, HAB, will be equiangular; for the angles HAB, KGH are right angles. Also the angles ABH, KHG are equal; wherefore, as KH (the excess of CK above AH) to GH, or to CA, the distance betwixt the first and second staff; so is AH, the distance betwixt the first and third staff, to AB the distance sought.

S C H O L I U M.

FIG. 10. To draw on a plane a right line AE perpendicular to CH, from a given point A; take the right lines AB, AD, on each side equal; and in the points B and D, let there be fixed stakes, to which let there be tied two equal ropes BE, DE, or one having a mark in the middle, and holding in your hand their extremities joined, (or the mark in the middle, if it be but one), draw out the ropes one the ground; and then, where the two ropes meet, or at the mark, when by it the rope is fully stretched, let there be placed a third stake at E; the right line AE will be perpendicular to CH in the point A (prob. 1. of part. 1.). In a manner not unlike to this, may any problems that are resolved by the square and compasses, be done by ropes and a cord turned round as a radius.

P R O P O S I T I O N IX.

FIG. 12. To measure the distance AB, one of whose extremities is accessible.—From the point A, let the right line AC of a known length be made perpendicular to AB, (by the preceding scholium): likewise draw the right line CD perpendicular to CB, meeting the right line AB in D: then as DA : AC :: AC : AB. Wherefore, when DA and AC are given, AB will be found by the rule of three.

S C H O L I U M.

All the preceding operations depend on the equality of some angles of triangles, and on the similarity of the triangles arising from that equality. And on the same principles

ciples depend innumerable other operations which a geometrical will find out of himself, as is very obvious. However, some of these operations require such exactness in the work, and without it are so liable to errors, that, ceteris paribus, the following operations, which are performed by a trigonometrical calculation, are to be preferred; yet could we not omit those above, being most easy in practice, and most clear and evident to those who have only the first elements of geometry. But if you are provided with instruments, the following operations are more to be relied upon. We do not insist on the easiest cases to those who are skilled in plain trigonometry, which is indeed necessary to any one who would apply himself to practice. See TRIGONOMETRY.

FIG. 13. To describe the construction and use of the geometrical quadrant.—The geometrical quadrant is the fourth part of a circle divided into ninety degrees, to which two sights are adapted, with a perpendicular or plumb line hanging from the centre. The general use of it is for investigating angles in a vertical plane, comprehended under right lines going from the centre of the instrument, one of which is horizontal, and the other is directed to some visible point. This instrument is made of any solid matter as wood, copper, &c.

FIG. 14. To describe and make use of the graphometer.—The graphometer is a semicircle made of any hard matter, of wood, for example, or brass, divided into 180 degrees; so fixed on a sulcrum, by means of a brass ball and socket, that it easily turns about, and retains any situation; two sights are fixed on its diameter. At the centre there is commonly a magnetical needle in a box. There is likewise a moveable ruler, which turns round the centre, and retains any situation given it. The use of it is to observe any angle, whose vertex is at the centre of the instrument in any plane, (though it is most commonly horizontal, or nearly so), and to find how many degrees it contains.

FIG. 15. and 16. To describe the manner in which angles are measured by a quadrant or graphometer.—Let there be an angle in a vertical plane, comprehended between a line parallel to the horizon HK, and the right line RA, coming from any remarkable point of a tower or hill, or from the sun, moon, or a star. Suppose that this angle RAH is to be measured by the quadrant: let the instrument be placed in the vertical plane, so as that the centre A may be in the angular point: and let the sights be directed towards the object at R, (by the help of the ray coming from it, if it be the sun or moon, or by the help of the visual ray, if it is anything else), the degrees and minutes in the arc BC cut off by the perpendicular, will measure the angle RAH required. For, from the make of the quadrant, BAD is a right angle; therefore BAR is likewise right, being equal to it. But, because HK is horizontal, and AC perpendicular, HAC will be a right angle; and therefore equal also to BAR. From those angles subtract the part HAB that is common to both; and there will remain the angle BAC equal to the angle RAH. But the arc BC is the measure of the

angle BAC; consequently, it is likewise the measure of the angle RAH.

Note. That the remaining arc on the quadrant DC is the measure of the angle RAZ, comprehended between the foresaid right line RA and AZ which points to the zenith.

Let it now be required to measure the angle ACB (fig. 16.) in any plane, comprehended between the right lines AC and BC, drawn from two points A and B, to the place of station C. Let the graphometer be placed at C, supported by its sulcrum (as was shown above); and let the immovable sights on the side of the instrument DE be directed towards the point A; and likewise (while the instrument remains immovable) let the sights of the ruler FG (which is moveable about the centre C) be directed to the point B. It is evident, that the moveable ruler cuts off an arc DH, which is the measure of the angle ACB sought. Moreover, by the same method, the inclination of CE, or of FG, may be observed with the meridian line, which is pointed out by the magnetic needle inclosed in the box, and is moveable about the centre of the instrument, and the measure of this inclination or angle found in degrees.

FIG. 17. To measure an accessible height by the geometrical quadrant.—By the 12th prop. of this part, let the angle C be found by means of the quadrant. Then in the triangle ABC, right angled at B, (BC being supposed the horizontal distance of the observator from the tower), having the angle at C, and the side BC, the required height BA will be found by the 3d case of plain trigonometry. See TRIGONOMETRY.

FIG. 18. To measure an inaccessible height by the geometrical quadrant.—Let the angle ACB be observed with the quadrant (by the 12th prop. of this part); then let the observer go from C to the second station D, in the right line BCD (provided BCD be a horizontal plane); and after measuring this distance CD, take the angle ADC likewise with the quadrant. Then, in the triangle ACD, there is given the angle ADC, with the angle ACD; because ACB was given before: therefore (by art. 59. of Part I.) the remaining angle CAD is given likewise. But the side CD is likewise given, being the distance of the station C and D; therefore (by the first case of oblique-angled triangles in trigonometry) the side AC will be found. Wherefore, in the right-angled triangle ABC, all the angles and the hypotenuse AC are given; consequently, by the fourth case of trigonometry, the height sought AB will be found; as also (if you please) the distance of the station C, from AB the perpendicular within the hill or inaccessible height.

FIG. 19. From the top of a given height, to measure the distance BC.—Let the angle BAC be observed by the 12th prop. of this; wherefore in the triangle ABC, right-angled at B, there is given by observation the angle at A; whence (by the 59th art. of Part I.) there will also be given the angle BCA: moreover the side AB (being the height of the tower) is supposed to be given. Wherefore,

Fig. 1.

A square grid with vertices labeled A, B, C, D. A diagonal line connects A and D. Points E, F, G, H are marked on the grid. A curved line is drawn from A towards the bottom right.

PlateXCV.

Diagram 3: A tower with a spire. A construction line connects the top of the spire to a point C on the ground. Points A, B, D are marked on the tower and ground respectively.

Diagram 2: A tall building. Construction lines connect the top of the building to points on the ground. Points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 4: A tower and a house. Construction lines connect the top of the tower to the top of the house. Points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 5: A small house and a church. Construction lines connect the top of the house to the top of the church. Points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 6: A building. A construction line connects the top of the building to a point on the ground. Points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 7: A tower. A construction line connects the top of the tower to a point on the ground. Points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 8: A building. A construction line connects the top of the building to a point on the ground. Points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 9: A geometric construction showing a triangle with vertices A, B, C and a point D. Points E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 11: A church. A construction line connects the top of the church to a point on the ground. Points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 12: A triangle with vertices A, B, C and a quarter-circle with center A. Points D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 10: A triangle with vertices A, B, C and a construction line. Points D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 15: A vertical line with construction points. Points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 16: A line with construction points. Points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 14: A protractor and a ruler. Points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 18: A church on a hill. Construction lines connect the top of the church to points on the ground. Points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 19: A building. A construction line connects the top of the building to a point on the ground. Points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 17: A tower. A construction line connects the top of the tower to a point on the ground. Points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 21: A landscape with a church and construction lines. Points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 22: A building on a hill. Construction lines connect the top of the building to points on the ground. Points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Diagram 20: A ship on the water. Construction lines connect the top of the ship to points on the ground. Points A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z are marked.

Wherefore, by the 3d case of trigonometry, BC, the distance sought, will be found.

PROPOSITION XVI.

FIG. 20. To measure the distance of two places A and B, of which one is accessible, by the graphometer.—Let there be erected at two points A and C, sufficiently distant, two visible signs; then (by the 12th prop. of this part) let the two angles BAC, BCA be taken by the graphometer. Let the distance of the stations A and C be measured with a chain. Then the third angle B being known, and the side AC being likewise known; therefore, by the first case of trigonometry, the distance required, AB, will be found.

PROPOSITION XVII.

FIG. 21. To measure by the graphometer, the distance of two places, neither of which is accessible.—Let two stations C and D be chosen, from each of which the places may be seen whose distance is sought: let the angles ACD, ACB, BCD, and likewise the angles BDC, BDA, CDA, be measured by the graphometer; let the distance of the stations C and D be measured by a chain, or (if it be necessary) by the preceding practice. Now, in the triangle ACD, there are given two angles ACD and ADC; therefore, the third CAD is likewise given; moreover the side CD is given; therefore, by the first case of trigonometry, the side AD will be found. After the same manner, in the triangle BCD, from all the angles and one side CD given, the side BD is found. Wherefore, in the triangle ADB, from the given sides DA and DB, and the angle ADB contained by them, the side AB (the distance sought) is found by the 4th case of trigonometry of oblique-angled triangles.

PROPOSITION XVIII.

FIG. 22. It is required by the graphometer and quadrant, to measure an accessible height AB, placed so on a steep, that one can neither go near it in an horizontal plane, nor recede from it, as we supposed in the solution of the 14th Prop.—Let there be chosen any situation as C, and another D; where let some mark be erected: let the angles ACD and ADC be found by the graphometer; then the third angle DAC will be known. Let the side CD, the distance of the stations, be measured with a chain, and thence (by trigon.) the side AC will be found. Again, in the triangle ACB, right-angled at B, having found by the quadrant the angle ACB, the other angle CAB is known likewise: but the side AC in the triangle ADC is already known; therefore the height required AB will be found by the 4th case of right-angled triangles. If the height of the tower is wanted, the angle BCF will be found by the quadrant; which being taken from the angle ACB already known, the angle ACF will remain: but the angle FAC was known before; therefore the remaining angle AFC will be known. But the side AC was also known before; therefore, in the triangle AFC, all the angles and one of the sides AC being known, AF, the height of the tower above the hill, will be found by trigonometry.

S C H O L I U M.

It were easy to add many other methods of measuring heights and distances; but, if what is above be understood, it will be easy (especially for one that is versed in

the elements) to contrive methods for this purpose, according to the occasion: so that there is no need of adding any more of this sort. We shall subjoin here a method by which the diameter of the earth may be found out.

PROPOSITION XIX.

PLATE CXVI. FIG. 1. To find the diameter of the earth from one observation.—Let there be chosen a high hill AB, near the sea-shore, and let the observator on the top of it, with an exact quadrant divided into minutes and seconds by transverse divisions, and fitted with a telescope in place of the common sights, measure the angle ABE contained under the right line AB, which goes to the centre, and the right line BE drawn to the sea, a tangent to the globe at E; let there be drawn from A perpendicular to BD, the line AF meeting BE in F. Now in the right-angled triangle BAF all the angles are given, also the side AB, the height of the hill; which is to be found by some of the foregoing methods, as exactly as possible; and (by trigonometry) the sides BF and AF are found. But, by cor. 36th 3. Eucl. AF is equal to FE; therefore BE will be known. Moreover, by 36th 3. Eucl. the rectangle under BA and BD is equal to the square of BE. And thence, by 17th 6. Eucl. as AB : BE :: BE : BD. Therefore, since AB and BE are already given, BD will be found by 11th, 6. Eucl. or by the rule of three; and subtracting BA, there will remain AD the diameter of the earth sought.

S C H O L I U M.

Many other methods might be proposed for measuring the diameter of the earth. The most exact is that proposed by Mr Picart of the academy of sciences at Paris.

“According to Mr Picart, a degree of the meridian at the latitude of 49° 21', was 57,060 French toises, each of which contains six feet of the same measure; from which it follows, that, if the earth be an exact sphere, the circumference of a great circle of it will be 123,249,600 Paris feet, and the semidiameter of the earth 19,615,800 feet: but the French mathematicians, who of late have examined Mr Picart's operations, assure us, That the degree in that latitude is 57,183 toises. They measured a degree in Lapland, in the latitude of 66° 20', and found it of 57,438 toises. By comparing these degrees, as well as by the observations on pendulums, and the theory of gravity, it appears that the earth is an oblate spheroid; and (supposing those degrees to be accurately measured) the axis or diameter that passes through the poles will be to the diameter of the equator, as 177 to 178, or the earth will be 22 miles higher at the equator than at the poles. A degree has likewise been measured at the equator, and found to be considerably less than at the latitude of Paris; which confirms the oblate figure of the earth. But an account of this last mensuration has not been published as yet. If the earth was of an uniform density from the surface to the centre, then, according to the theory of gravity, the meridian would be an exact ellipsis, and the axis would be to the diameter of the equator as 230 to 231; and the difference of the semidiameter of the equator and semiaxis about 17 miles.”

In what follows, a figure is often to be laid down on paper, like to another figure given; and because this likeness consists in the equality of their angles, and in the sides having the same proportion to each other (by the definitions of the 6th of Eucl.) we are now to shew what methods practical geometers use for making on paper an angle equal to a given angle, and how they constitute the sides in the same proportion. For this purpose they make use of a protractor, (or, when it is wanting, a line of chords), and of a line of equal parts.

PROPOSITION XX.

FIG. 2. 3. 4. 5. and 6. To describe the construction and use of the protractor, of the line of chords, and of the line of equal parts. The protractor is a small semicircle of brass, or such solid matter. The semicircumference is divided into 180 degrees. The use of it is, to draw angles on any plane, as on paper, or to examine the extent of angles already laid down. For this last purpose, let the small point in the centre of the protractor be placed above the angular point, and let the side AB coincide with one of the sides that contain the angle proposed; the number of degrees cut off by the other side, computing on the protractor from B, will show the quantity of the angle that is to be measured.

But if an angle is to be made of a given quantity on a given line, and at a given point of that line, let AB coincide with the given line, and let the centre A of the instrument be applied to that point. Then let there be a mark made at the given number of degrees; and a right line drawn from that mark to the given point, will constitute an angle with the given right line of the quantity required; as is manifest.

This is the most natural and easy method, either for the extent of an angle on paper, or for describing on paper an angle of a given quantity.

But when there is scarcity of instruments, or because a line of chords is more easily carried about, (being described on a ruler on which there are many other lines besides), practical geometers frequently make use of it. It is made thus: let the quadrant of a circle be divided into 90 degrees; (as in fig. 4.) The line AB is the chord of 90 degrees; the chord of every arc of the quadrant is transferred to this line AB, which is always marked with the number of degrees in the corresponding arc.

Note, that the chord of 60 degrees is equal to the radius, by corol. 15. 4th Eucl. If now a given angle EDF is to be measured by the line of chords from the centre D, with the distance DG, (the chord of 60 degrees,) describe the arch GF; and let the points G and F be marked where this arch intersects the sides of the angle. Then if the distance GF, applied on the line of chords from A to B, gives (for example) 25 degrees, this shall be the measure of the angle proposed.

When an obtuse angle is to be measured with this line, let its complement to a semicircle be measured, and thence it will be known. It were easy to transfer to the diameter of a circle the chords of all arches to the extent of a semicircle; but such are rarely found marked upon rules.

But now, if an angle of a given quantity, suppose of 50 degrees, is to be made at a given point M of the right line KL (fig. 6.) From the centre M, and the di-

stance MN, equal to the chord of 60 degrees, describe the arc QN. Take off an arc NR, whose chord is equal to that of 50 degrees on the line of chords; join the points M and R; and it is plain that MR shall contain an angle of 50 degrees with the line KL proposed.

But sometimes we cannot produce the sides, till they be of the length of a chord of 60 degrees on our scale; in which case it is fit to work by a circle of proportions (that is a sector); by which an arc may be made of a given number of degrees to any radius.

The quantities of angles are likewise determined by other lines usually marked upon rules, as the lines of sines, tangents, and secants; but, as these methods are not so easy or so proper in this place, we omit them.

To delineate figures similar or like to others given, besides the equality of the angles, the same proportion is to be preserved among the sides of the figure that is to be delineated, as is among the sides of the figures given. For which purpose, on the rules used by artists, there is a line divided into equal parts, more or less in number, and greater or lesser in quantity, according to the pleasure of the maker.

A foot is divided into inches; and an inch, by means of transverse lines, into 100 equal parts: so that with this scale, any number of inches, below twelve, with any part of an inch, can be taken by the compasses, providing such part be greater than the one hundredth part of an inch. And this exactness is very necessary in delineating the plans of houses, and in other cases.

PROPOSITION XXI.

FIG. 7. To lay down on paper, by the protractor or line of chords, and line of equal parts, a right-lined figure like to one given, providing the angles and sides of the figure given be known by observation or mensuration. —For example, suppose that it is known that in a quadrangular figure, one side is of 235 feet, that the angle contained by it and the second side is of 84°, the second side of 288 feet, the angle contained by it and the third side of 72°, and that the third side is 294 feet. These things being given, a figure is to be drawn on paper like to this quadrangular figure. On your paper, at a proper point A, let a right line be drawn, upon which take 235 equal parts, as AB. The part representing a foot is taken greater or lesser, according as you would have your figure greater or less. In the adjoining figure, the 100th part of an inch is taken for a foot. And accordingly an inch divided into 100 parts, and annexed to the figure, is called a scale of 100 feet. Let there be made at the point B (by the preceding prop.) an angle ABC of 85°, and let BC be taken of 288 parts like to the former. Then let the angle BCD be made of 72°, and the side CD of 294 equal parts. Then let the side AD be drawn; and it will complete the figure like to the figure given. The measures of the angle A and D can be known by the protractor or line of chords, and the side AD by the line of equal parts; which will exactly answer to the corresponding angles and to the side of the primary figure.

After the very same manner, from the sides and angles given, which bound any right-lined figure, a figure like to it may be drawn, and the rest of its sides and angles be known.

Hence any trigonometrical problem in right-lined triangles, may be resolved by delineating the triangle from what is given concerning it, as in this proposition. The unknown sides are examined by a line of equal parts, and the angles by a protractor or line of chords.

The diameter of a circle being given, to find its circumference nearly.—The periphery of any polygon inscribed in the circle is less than the circumference, and the periphery of any polygon described about a circle is greater than the circumference. Whence Archimedes first discovered that the diameter was in proportion to the circumference, as 7 to 22 nearly; which serves for common use. But the moderns have computed the proportion of the diameter to the circumference to greater exactness. Supposing the diameter 100, the periphery will be more than 314, but less than 315. The diameter is more nearly to the circumference, as 113 to 355. But Ludolphus van Cuelen exceeded the labours of all; for by immense study he found, that, supposing the diameter 100,000,000,000,000,000,000,000,000,000,000, the periphery will be less than 314,159,265,358,979,323,846,264,338,327,951, but greater than 314,159,265,358,979,323,846,264,338,327,950; whence it will be easy, any part of the circumference being given in degrees and minutes, to assign it in parts of the diameter.

HITHERTO we have treated of the measuring of angles and sides, whence it is abundantly easy to lay down a field, a plane, or an entire country: for to this nothing is requisite but the protraction of triangles, and of other plain figures, after having measured their sides and angles. But as this is esteemed an important part of practical geometry, we shall subjoin here an account of it with all possible brevity; suggesting withal, that a surveyor will improve himself more by one day's practice, than by a great deal of reading.

To explain what surveying is, and what instruments Surveyors use.—First, it is necessary that the surveyor view the field that is to be measured, and investigate its sides and angles, by means of an iron chain (having a particular mark at each foot of length, or at any number of feet, as may be most convenient for reducing lines or surfaces to the received measures), and the graphometer described above. Secondly, It is necessary to delineate the field in plano, or to form a map of it; that is, to lay down on paper a figure similar to the field; which is done by the protractor (or line of chords) and of the line of equal parts. Thirdly, It is necessary to find out the area of the field so surveyed and represented by a map. Of this last we are to treat below.

The sides and angles of small fields are surveyed by the help of a plain table; which is generally of an oblong rectangular figure, and supported by a sulcrum, so as to turn every way by means of a ball and socket. It has a

moveable frame, which surrounds the board, and serves to keep a clean paper put on the board close and tight to it. The sides of the frame facing the paper are divided into equal parts every way. The board hath besides a box with a magnetic needle, and moreover a large index with two sights. On the edge of the frame of the board are marked degrees and minutes, so as to supply the room of a graphometer.

FIG. 8. To delineate a field by the help of a plain-table, from one station whence all its angles may be seen, and their distances measured by a chain.—Let the field that is to be laid down be ABCDE. At any convenient place F, let the plain-table be erected; cover it with clean paper, in which let some point near the middle represent the station. Then applying at this place the index with the sights, direct it so as that through the sights some mark may be seen at one of the angles, suppose A; and from the point F, representing the station, draw a faint right line along the side of the index: then, by the help of the chain, let FA the distance of the station from the foresaid angle be measured. Then taking what part you think convenient for a foot or pace from the line of equal parts, set off on the faint line the parts corresponding to the line FA that was measured; and let there be a mark made representing the angle of the field A. Keeping the table immovable, the same is to be done with the rest of the angles; then right lines joining those marks shall include a figure like to the field, as is evident from 5. 6. Eucl.

The same thing is done in like manner by the graphometer: for having observed in each of the triangles, AFB, BFC, CFD, &c. the angle at the station F, and having measured the lines from the station to the angles of the field, let similar triangles be protracted on paper (by the 21. prop. of this) having their common vertex in the point of station. All the lines, excepting those which represent the sides of the field, are to be drawn faint or obscure.

Note 1. When a surveyor wants to lay down a field, let him place distinctly in a register all the observations of the angles, and the measures of the sides, until, at time and place convenient, he draw out the figure on paper.

Note 2. The observations made by the help of the graphometer are to be examined: for all the angles about the point F ought to be equal to four right ones. (by cor. 2. art. 30. of part I.)

FIG. 9. To lay down a field by means of two stations, from each of which all the angles can be seen, by measuring only the distance of the stations.—Let the instrument be placed at the station F: and having chosen a point representing it upon the paper which is laid upon the plain table, let the index be applied at this point, so as to be moveable about it. Then let it be directed successively to the several angles of the field: and when any angle is seen through the sights, draw an obscure line along the side of the index. Let the index, with the sights, be directed after the same manner to the station C: on the obscure line drawn along its side, pointing to A,

A, set off from the scale of equal parts a line corresponding to the measured distance of the stations, and this will determine the point G. Then remove the instrument to the station G, and applying the index to the line representing the distance of the stations, place the instrument so that the first station may be seen through the sights. Then the instrument remaining immovable, let the index be applied at the point representing the second station G, and be successively directed by means of its sights, to all the angles of the field, drawing (as before) obscure lines: and the intersection of the two obscure lines that were drawn to the same angle from the two stations will always represent that angle on the plan. Care must be taken that those lines be not mistaken for one another. Lines joining those intersections will form a figure on the paper like to the field.

S C H O L I U M.

It will not be difficult to do the same by the graphometer, if you keep a distinct account of your observations of the angles made by the line joining the stations, and the lines drawn from the stations to the respective angles of the field. And this is the most common manner of laying down whole countries. The tops of two mountains are taken for two stations, and their distance is either measured by some of the methods mentioned above, or is taken according to common repute. The sights are successively directed towards cities, churches, villages, forts, lakes, turnings of rivers, woods, &c.

Note, The distance of the stations ought to be great enough, with respect to the field that is to be measured; such ought to be chosen as are not in a line with any angle of the field. And care ought to be taken likewise that the angles, for example, FAG, FDG, &c. be neither very acute, nor very obtuse. Such angles are to be avoided as much as possible; and this admonition is found very useful in practice.

P R O P O S I T I O N XXVI.

FIG. 10. To lay down any field, however irregular its figure may be, by the help of the graphometer.—Let ABCEDHG be such a field. Let its angles (in going round it) be observed with a graphometer (by the 12. of this) and noted down; let its sides be measured with a chain; and (by what was said on the 21. of this) let a figure like to the given field be protracted on paper. If any mountain is in the circumference, the horizontal line hid under it is to be taken for a side, which may be found by two or three observations according to some of the methods described above; and its place on the map is to be distinguished by a shade, that it may be known a mountain is there.

If not only the circumference of the field is to be laid down on the plan, but also its contents, as villages, gardens, churches, public roads, we must proceed in this manner.

Let there be (for example) a church F, to be laid down in the plan. Let the angles ABF BAF be observed and protracted on paper in their proper places, the intersection of the two sides BF and AF will give the place of the church on the paper: Or, more exactly, the lines BF AF being measured, let circles be described from the centres B and A, with parts from the

scale corresponding to the distances BF and AF, and the place of the church will be at their intersection.

Note 1. While the angles observed by the graphometer are taken down, you must be careful to distinguish the external angles, as E and G, that they may be rightly protracted afterwards on paper.

Note 2. Our observations of the angles may be examined by computing if all the internal angles make twice as many right angles, four excepted, as there are sides of the figure: (for this is demonstrated by 32. 1. Eucl.) But in place of any external angle DEC, its complement to a circle is to be taken.

P R O P O S I T I O N XXVII.

FIG. 11. To lay down a plain field without instruments.—If a small field is to be measured, and a map of it to be made, and you are not provided with instruments; let it be supposed to be divided into triangles, by right-lines, as in the figure; and after measuring the three sides of any of the triangles, for example of ABC, let its sides be laid down from a convenient scale on paper, (by the 22. of this.) Again, let the other two sides BD CD of the triangle CBD be measured and protracted on the paper by the same scale as before. In the same manner proceed with the rest of the triangles of which the field is composed, and the map of the field will be perfected; for the three sides of a triangle determine the triangle; whence each triangle on the paper is similar to its correspondent triangle in the field, and is similarly situated; consequently the whole figure is like to the whole field.

S C H O L I U M.

If the field be small, and all its angles may be seen from one station, it may be very well laid down by the plain-table, (by the 24. of this.) If the field be larger, and have the requisite conditions, and great exactness is not expected, it likewise may be plotted by means of the plain-table, or by the graphometer, (according to the 25. of this; but in fields that are irregular and mountainous, when an exact map is required, we are to make use of the graphometer, (as in the 26. of this,) but rarely of the plain-table.

Having protracted the bounding lines, the particular parts contained within them may be laid down by the proper operations for this purpose, (delivered in the 26th proposition; and the method described in the 27th proposition may be sometimes of service;) for we may trust more to the measuring of sides, than to the observing of angles. We are not to compute four-sided and many sided figures till they are resolved into triangles: for the sides do not determine those figures.

In the laying down of cities, or, the like, we may make use of any of the methods described above that may be most convenient.

The map being finished, it is transferred on clean paper, by putting the first sketch above it, and marking the angles by the point of a small needle. These points being joined by right lines, and the whole illuminated by colours proper to each part, and the figure of the mariners compass being added to distinguish the north and south, with a scale on the margin, the map or plan will be finished and neat.

We have thus briefly and plainly treated of surveying, and shown by what instruments it is performed; having avoided those methods which depend on the magnetic needle, not only because its direction may vary in different places of a field (the contrary of this at least doth not appear,) but because the quantity of an angle observed by it cannot be exactly known; for an error of two or three degrees can scarcely be avoided in taking angles by it.

As for the remaining part of surveying, whereby the area of a field already laid down on paper is found in acres, roods, or any other superficial measures; this we leave to the following section, which treats of the mensuration of surfaces.

Besides the instruments described above, a surveyor ought to be provided with an off-set staff equal in length to ten links of the chain, and divided into ten equal parts. He ought likewise to have ten arrows or small straight sticks near two feet long, shod with iron ferrils. When the chain is first opened, it ought to be examined by the off-set staff. In measuring any line, the leader of the chain is to have the ten arrows at first setting out. When the chain is stretched in the line, and the near end touches the place from which you measure, the leader sticks one of the ten arrows in the ground, at the far end of the chain. Then the leader leaving the arrow, proceeds with the chain another length; and the chain being stretched in the line, so that the near end touches the first arrow, the leader sticks down another arrow at his end of the chain. The line is preserved straight, if the arrows be always set so as to be in a right line with the place you measure from, and that to which you are going. In this manner they proceed till the leader have no more arrows. At the eleventh chain, the arrows are to be carried to him again, and he is to stick one of them into the ground, at the end of the chain. And the same is to be done at the 21, 31, 41, &c. chains, if there are so many in a right line to be measured. In this manner you can hardly commit an error in numbering the chains, unless of ten chains at once.

The off-set staff serves for measuring readily the distances of any things proper to be represented in your plan, from the station-line while you go along. These distances ought to be entered into your field-book, with the corresponding distances from the last station, and proper remarks, that you may be enabled to plot them justly, and be in no danger of mistaking one for another when you extend your plan. The field-book may be conveniently divided into five columns. In the middle column the angles at the several stations taken by the theodolite are to be entered, with distances from the stations. The distances taken by the off-set staff, on either side of the station-line, are to be entered into columns on either side of the middle column, according to their position with respect to that line. The names and characters of the objects, with proper remarks, may be entered in columns on either side of these last.

Because, in the place of the graphometer described by our author, surveyors now make use of the theodolite, we shall subjoin a description of Mr Sisson's latest

improved theodolite from Mr Gardner's practical surveying improved. See a figure of it in Plate XCVI.

In this instrument, the three staves, by brass ferrils at top, screw into bell-metal joints, that are moveable between brass pillars, fixed in a strong brass plate; in which, round the centre, is fixed a socket with a ball moveable in it, and upon which the four screws press, that set the limb horizontal: Next above is another such plate, through which the said screws pass, and on which, round the centre, is fixed a frustum of a cone of bell-metal, whose axis (being connected with the centre of the bell) is always perpendicular to the limb, by means of a conical brass ferril fitted to it, whereon is fixed the compass-box; and on it the limb, which is a strong bell-metal ring, whereon are moveable three brass indexes; in whose plate are fixed four brass pillars, that, joining at top, hold the centre pin of the bell-metal double sextant, whose double index is fixed on the centre of the same plate: Within the double sextant is fixed the spirit-level, and over it the telescope.

The compass box is graven with two diamonds for north and south, and with 20 degrees on both sides of each, that the needle may be set to the variation, and its error also known.

The limb has two flours de luce against the diamonds in the box, instead of 180 each; and is curiously divided into whole degrees, and numbered to the left hand at every ten to twice 180, having three indexes distant 120, (with Nonius's divisions on each for the decimals of a degree), that are moved by a pinion fixed below one of them, without moving the limb; and in another is a screw and spring under, to fix it to any part of the limb. It has also divisions numbered, for taking the quarter girt in inches of round timber at the middle height, when standing ten feet horizontally distant from its centre; which at 20 must be doubled, and at 30 tripled; to which a shorter index is used, having Nonius's divisions for the decimals of an inch; but an abatement must be made for the bark, if not taken off.

The double sextant is divided on one side from under its centre (when the spirit-tube and telescope are level) to above 60 degrees each way, and numbered at 10, 20, &c. and the double index (through which it is moveable) shews on the same side the degree and decimal of any altitude or depression to that extent by Nonius's divisions: On the other side are divisions numbered, for taking the upright height of timber, &c. in feet, when distant 10 feet; which at 20 must be doubled, and at 30 tripled; and also the quantities for reducing hypothenusal lines to horizontal. It is moveable by a pinion fixed in the double index.

The telescope is a little shorter than the diameter of the limb, that a fall may not hurt it; yet it will magnify as much, and shew a distant object as perfect, as most of triple its length. In its focus are very fine cross wires, whose intersection is in the plane of the double sextant; and this was a whole circle, and turned in a lathe to a true plane, and is fixed at right angles to the limb; so that, whenever the limb is set horizontal, (which is readily done by making the spirit-tube level

“ level over two screws, and the like over the other two),
“ the double sextant and telescope are moveable in a ver-
“ tical plane; and then every angle taken on the limb
“ (though the telescope be never so much elevated or
“ depressed) will be an angle in the plane of the horizon.
“ And this is absolutely necessary in plotting a horizontal
“ plane.

“ If the lands to be plotted are hilly, and not in any
“ one plane, the lines measured cannot be truly laid down
“ on paper, without being reduced to one plane, which
“ must be the horizontal, because angles are taken in that
“ plane.—

“ In viewing your objects, if they have much altitude
“ or depression, either write down the degree and decimal
“ shewn on the double sextant, or the links shewn on the
“ back side; which last subtracted from every chain in
“ the station-line, leaves the length in the horizontal
“ plane. But if the degree is taken, the following table
“ will shew the quantity.

A table of the links to be subtracted out of every chain
in hypothensal lines of several degrees altitude, or
depression, for reducing them to horizontal.

“ Let the first station-line really measure 1107 links,
“ and the angle of altitude or depression be 19°, 95;
“ looking in the table you will find against 19°, 95, is 6
“ links. Now 6 times 11 is 66; which subtracted from
“ 1107, leaves 1041, the true length to be laid down
“ in the plan.

“ It is useful in surveying, to take the angles, which
“ the bounding lines form, with the magnetic needle,
“ in order to check the angles of the figure, and to plot
“ them conveniently afterwards.”

Of the Surfaces of Bodies.

THE smallest superficial measure with us is a square
inch; 144 of which make a square foot. Wrights make
use of these in the measuring of deals and planks; but
the square foot which the glaziers use in measuring of
glass, consists only of 64 square inches. The other
measures are, first, the ell square; secondly, the fall,
containing 36 square ells; thirdly, the rood, containing
40 falls; fourthly, the acre, containing 4 roods. Slaters,
masons, and paviours, use the ell square and the fall; sur-
veyors of land use the square ell, the fall, the rood, and
the acre.

The superficial measures of the English are, first, the
square foot; secondly, the square yard, containing 9
square feet, for their yard contains only 3 feet; thirdly,
the pole, containing 30$\backslash frac\{1\}\{2\}$ square yards; fourthly, the
rood, containing 40 poles; fifthly, the acre, containing

4 roods. And hence it is easy to reduce our superficial
measures to the English, or theirs to ours.

“ In order to find the content of a field, it is most con-
“ venient to measure the lines by the chains described a-
“ bove, p. 693, that of 22 yards for computing the Eng-
“ lish acres, and that of 24 Scots ells for the acres of
“ Scotland. The chain is divided into 100 links, and the
“ square of the chain is 10,000 square links; ten squares
“ of the chain, or 100,000 square links, give an acre.
“ Therefore, if the area be expressed by square links,
“ divide by 100,000, or cut off five decimal places, and
“ the quotient shall give the area in acres and decimals
“ of an acre. Write the entire acres apart; but multi-
“ ply the decimals of an acre by 4, and the product shall
“ give the remainder of the area in roods and decimals
“ of a rood. Let the entire roods be noted apart after
“ the acres; then multiply the decimals of a rood by 40,
“ and the product shall give the remainder of the area in
“ falls or poles. Let the entire falls or poles be then
“ writ after the roods, and multiply the decimals of a
“ fall by 36, if the area is required in the measures of
“ Scotland; but multiply the decimals of a pole by 30$\backslash frac\{1\}\{2\}$,
“ if the area is required in the measures of England, and
“ the product shall give the remainder of the area in
“ square ells in the former case, but in square yards in
“ the latter. If, in the former case, you would reduce
“ the decimals of the square ell to square feet, multiply
“ them by 9,50994; but, in the latter case, the decimals
“ of the English square yard are reduce to square feet,
“ by multiplying them by 9.

“ Suppose, for example, that the area appears to con-
“ tain 12,65842 square links of the chain of 24 ells;
“ and that this area is to be expressed in acres, roods,
“ falls, &c. of the measures of Scotland. Divide the
“ square links by 100,000, and the quotient 12,65842
“ shows the area to contain 12 acres $\backslash frac\{65842\}\{100000\}$ of an acre.
“ Multiply the decimal part by 4, and the product
“ 263368 gives the remainder in roods and decimals of
“ a rood. Those decimals of the rood being multiplied
“ by 40, the product gives 25,3472 falls. Multiply the
“ decimals of the fall by 36, and the product gives
“ 12,4992 square ells. The decimals of the square ell
“ multiplied by 9,50994 give 4,7458 square feet.
“ Therefore the area proposed amounts to 12 acres, 2
“ roods, 25 falls, 12 square ells, and 4,7458 square
“ feet.

“ But if the area contains the same number of square
“ links of Gunter's chain, and is to be expressed by
“ English measures, the acres and roods are computed in
“ the same manner as in the former case. The poles are
“ computed as the falls. But the decimals of the pole,
“ viz. $\backslash frac\{1\}\{30\backslash frac\{1\}\{2\}\}$, are to be multiplied by 30$\backslash frac\{1\}\{2\}$ (or 30.25),
“ and the product gives 10,5028 square yards. The de-
“ cimals of the square yard, multiplied by 9, give 4,5252
“ square feet; therefore, in this case, the area is in
“ English measure 12 acres, 2 roods, 25 poles, 10 square
“ yards, and 4,5252 square feet.

“ The Scots acre is to the English acre, by statute,
“ as 100,000 to 78,694, if we have regard to the diffe-
“ rence betwixt the Scots and English foot above men-
“ tioned. But it is customary in some parts of England
“ to

" to have 18, 21 &c. feet to a pole, and 160 such poles
" to an acre; whereas, by the statute, 16$\backslash frac\{1\}\{2\}$ feet make a
" pole. In such cases the acre is greater in the dupli-
" cate ratio of the number of feet to a pole.

" They who measure land in Scotland by an ell of 37
" English inches, make the acre less than the true Scots
" acre by 593$\backslash frac\{1\}\{2\}$ square English feet, or by about $\backslash frac\{1\}\{3\}$ of
" the acre.

" An husband-land contains 6 acres of fock and sythe-
" land, that is, of land that may be tilled with a plough,
" and mown with a sythe; 13 acres of arable land make
" an oxgang or oxengate; four oxengate make a pound-
" land of old extent (by a decree of the Exchequer,
" March 11. 1585), and is called librata terre. A
" forty-shilling land of old extent contains eight oxgang,
" or 104 acres.

" The arpent, about Paris, contains 32400 square Pa-
" ris feet, and is equal to 2$\backslash frac\{1\}\{2\}$ Scots roods, or 7$\backslash frac\{1\}\{2\}$ Eng-
" lish roods.

" The astus quadratus, according to Varro, Collu-
" mella, &c. was a square of 120 Roman feet. The
" jugerum was the double of this. It is to the Scots a-
" cre as 10,000 to 20,456, and to the English acre as
" 10,000 to 16,097. It was divided (like the as) into
" 12 uncie, and the uncia into 24 scrupula.—This,
" with the three preceding paragraphs, are taken from an
" ingenious manuscript, written by Sir Robert Stewart
" professor of natural philosophy. The greatest part of
" the table in p. 693. was taken from it likewise.

PROPOSITION XXVIII.

FIG. 12. To find out the area of a rectangular paral-
lelogram 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 1.

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 mul-
tiplied into the base, or the half of the base into the per-
pendicular, 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 shown in the 31st prop.

PROPOSITION XXX.

FIG. 14. To find the area of any rectilinear 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.

SCHOLIUM 1.

In measuring boards, planks, and glass, 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 parallelo-
grams.

SCHOLIUM 2.

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 3.

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 mea-
sured on the uneven surface of the field. And indeed it
would appear, that the horizontal plane is to be account-
ed 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 stand perpendicular to
the horizon, it is plain, that a mountainous country can-
not be considered as of greater extent for 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 ex-
ceeds the area of the horizontal figure more or less, ac-
cording as the field is more or less uneven.

PROPOSITION XXXI.

FIG. 13. The sides of a triangle being given, to find the a-
rea, without finding the perpendicular.—Let all the sides of
the triangle be collected into one sum; from the half of
which let the sides be separately subtracted, that three
differences may be found betwixt the foresaid half sum
and each side; then let these three differences and the
half sum be multiplied into one another, and the square
root of the product will give the area of the triangle. For
example, let the sides be 10, 17, 21; the half of their
sum is 24; the three differences betwixt this half sum
and the three sides, are 14, 7, and 3. The first being
multiplied by the second, and their product by the third,
we have 294 for the product of the differences; which
multiplied by the foresaid half sum 24, gives 7056; the
square root of which 84 is the area of the triangle. The
demonstration of this, for the sake of brevity, we omit.
It is to be found in several treatises, particularly in Cla-
vius's Practical Geometry.

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 res-
olved 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.

C O R O L L A R Y.

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 squares 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 2d 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 $38\backslash frac\{1\}\{2\}$; therefore the square DC will be to the inscribed circle as 49 to $38\backslash frac\{1\}\{2\}$, 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 $1-\backslash frac\{1\}\{2\}+\backslash frac\{1\}\{4\}-\backslash frac\{1\}\{8\}+\backslash frac\{1\}\{16\}-\backslash frac\{1\}\{32\}+\backslash frac\{1\}\{64\}$, &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.

S C H O L I U M.

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 rectilinear 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 tetrahedron consists of four equal and equiangular triangles; the superficies of the hexahedron, or cube, of six equal squares; an octahedron, of eight equal equilateral triangles; a dodecahedron, of twelve equal and ordinate pentagons; and the superficies of an icosahedron, 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.

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 basis) 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.

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.

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

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.

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 sphaera) 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.

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.

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.

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 firlot, by statute, contains 10½ 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 $\backslash frac\{1\}\{2\}$ $\backslash frac\{1\}\{2\}$ $\backslash frac\{1\}\{2\}$ cubical inches), or any vessel containing 231 cubical inches, and no more, is deemed to be a lawful wine-gallon. An English pint therefore contains 28½ cubical inches; two pints make a quart; four quarts a gallon; 18 gallons a roundlet; three roundlets and an half, or 63 gallons, make a hogshhead; the half of a hogshhead is a barrel: one hogshhead and a third, or 84 gallons, make a puncheon; one puncheon and a half, or two hogshheads, 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 hogshheads, make a ton of wine. Though the English wine gallon is now fixed at 231 cubical inches, the standard kept in Guildhall being measured, before many persons of distinction, May 25th 1688, it was found to contain only 224 such inches.

“In the English beer measure, a gallon contains 282 cubical inches; consequently 35½ cubical inches make a pint, two pints make a quart, four quarts make a

"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 cubical 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 Edinburgh, marked upon it) having been carefully measured several times, and by different persons, the Scots pint, according to those standards, was found to contain about $103\backslash frac\{1\}\{2\}$ cubic inches. The pewterers jugs (by which the vessels in common use are made) are said to contain sometimes between 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 $46\backslash frac\{1\}\{2\}$ Scots pints; the same vessel contained $18\backslash frac\{1\}\{2\}$ English ale gallons. Supposing this mensuring 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 suspected, on several grounds, that the experiment was not made with sufficient care and exactness.

"The commissioners appointed by authority of parliament to settle the measures and weights, in their act of February 19. 1618, relate, That having caused fill the Linlithgow firlet with water, they found that it contained $21\backslash frac\{1\}\{2\}$ pints of the just Stirling jug and measure. They likewise ordain that this shall be the just and only firlet; and add, That the wideness and breadness of the which firlet, under and above even over within the bairds, shall contain nineteen inches and the sixth 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 content of such a vessel as is described in the act, is $2115.85$, and this divided by $21\backslash frac\{1\}\{2\}$ gives 99.56. But, by the weight of water said to fill this firlet in the same act, the measure of the pint agrees nearly with the Edinburgh standard above mentioned.

"As for the English measures of corn, the Winchester gallon contains $272\backslash frac\{1\}\{2\}$ 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 $19\backslash frac\{1\}\{2\}$ Scots pints make a firlet. 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 firlet. 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 $19\backslash frac\{1\}\{2\}$ such pints will amount to 2118.94, cubical inches; which agrees nearly with $2115.85$, the measure of the firlet by statute above mentioned. But it is probable, that in this he followed the act 1587, where it is ordained, That the wheat-firlet shall contain 19 pints and two joucattes. A wheat-firlet marked with the Linlithgow stamps being measured, was found to contain about 2211 cubical inches. By the statute of 1618 the barley-firlet 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 Beiffan contains 644 68099 Paris cubical inches, or 780 36 English cubical inches.

"The Roman amphora was a cubical Roman foot, the congius was the eighth part of the amphora, the sextarius was one sixth of the congius. They divided the sextarius like the as or libra. Of dry measures, the medius was equal to two amphoras, that is, about $1\backslash frac\{1\}\{2\}$ 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 pyramid, or the third part of the base by the height, the product will give the solid content, by 7th 12. Eucl.

COROLLARY.

If the solid content of a frustum of a pyramid is required, 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 cylinder, as ABCD. If towards the middle EF it be somewhat grosser, 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 diameters 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

area of the base (found by prop. 33.) be multiplied into $\backslash frac\{1\}\{3\}$ 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.

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 scalenous, 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 22d 5. Eucl.) as CG:AH::CD:EF; 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.

FIG. 26. Some cainks whose staves are remarkably bended about the middle, and strait towards the ends may be taken for two portions of cones, without any considerable error. Thus ABEG 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.

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.) $EAq\; =\; HFq\; =\; EFq$ 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 every where, 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 $\backslash frac\{2\}\{3\}$ 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 inferred to be that of 3 to 2.

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.

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.

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.

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 every where, 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.

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.

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 $CF\_9$ to $CD\_9$.

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.

C O R O L L A R Y 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.

C O R O L L A R Y 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 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.

P R O P O S I T I O N 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.

P R O P O S I T I O N LII.

To find the solid content of a regular and ordinate

body.—A tetrahedron being a pyramid, the solid content is found by the 44th prop. The hexahedron, or cube, being a kind of prism, it is measured by the 43d prop. An octahedron consists of two pyramids of the same square base and of equal heights; consequently its measure is found by the 44th prop. A dodecahedron consists of twelve pyramids having equal equilateral and equiangular pentagonal bases; and so 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 dodecahedron. The icosahedron 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.

P R O P O S I T I O N 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 $DDH\; \backslash times\; .0034$, or $DDH\; \backslash times\; .0034$. Ex. Let the diameter $D\; =\; 51.2$ inches, the height $H\; =\; 62.3$ inches, then the content shall be $51.2\; \backslash times\; 51.2\; \backslash times\; 62.3\; \backslash times\; .0034\; =\; 555.27.342$ 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 $DD\; \backslash times\; .7854$; and by the latter, the content of the vessel in solid inches

Isen's Theodolite

Fig. 1.

" inches is $DDHX.7854$; which divided by 231 (the
" number of cubical inches in a wine-gallon) gives $DDH$
" $\backslash times\; .0034$, 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 sale, 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 com-
" puted 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 pro-
" duct 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 com-
" puted in those bushels by multiplying the square of the
" diameter of the vessel by the height, and the product
" by the decimal fraction $.0,003,606$. But the stan-
" dard 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 $18\backslash frac\{1\}\{2\}$ inches dia-
" meter throughout, and 8 inches deep, should be esteem-
" ed 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 diam-
" eter 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 $21\backslash frac\{1\}\{2\}$
" 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 firlets is computed by multiplying the square of
" the diameter by the height, and their product by the
" decimal fraction $.00,358$. This firlet, in 1426, is
" appointed to contain 17 pints; in 1457, it was appoint-
" ed to contain 18 pints; in 1587, it is $19\backslash frac\{1\}\{2\}$ pints; in
" 1628, it is $21\backslash frac\{1\}\{2\}$ pints: and though this last statute ap-
" pears 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 firlet seems to be the
" least exceptionable; and therefore we suppose the fir-
" lot to contain $21\backslash frac\{1\}\{2\}$ pints of the Edinburgh standard, or
" about 2197 cubical inches; which a little exceeds the
" Winchester bushel, from which it may have been ori-
" ginally copied.

" VI. Supposing the bear-firlet 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 firlets 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 diam-
" eter 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 ex-
" ample, the content in wine-gallons is $AA\; \backslash times\; AB\; \backslash times\; BB$
" $\backslash times\; \backslash frac\{1\}\{3\}\; \backslash times\; H\; \backslash times\; .0034$.

" 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 $\backslash frac\{1\}\{2\}\; A\; \backslash times\; \backslash frac\{1\}\{2\}\; B$ added to $\backslash frac\{1\}\{3\}$ of the square of $\backslash frac\{1\}\{2\}\; A\; -\; \backslash frac\{1\}\{2\}\; B$, gives $\backslash frac\{1\}\{3\}\; AA\; \backslash times\; \backslash frac\{1\}\{3\}\; AB\; \backslash times\; \backslash frac\{1\}\{3\}\; BB$.

" VIII. When a vessel is a frustum of a parabolic
" conoid, measure the diameter of the section at the mid-
" dle 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 mea-
" sure 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 $\backslash frac\{1\}\{3\}$ 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 con-
" tent of the vessel. That is, if D represent the diam-
" eter of the middle section, and H the height of the
" frustum, you are to substitute $DD\; -\; \backslash frac\{1\}\{3\}\; HH$ 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 dia-
" meter of the middle section of the frustum, H the
" height or length of the frustum; and substitute in the
" first six rules $DD\; -\; \backslash frac\{HH\}\{3n^2\}$ for the square of the square
" of the diameter of the vessel.

" XI. When the vessel is a 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 $DD\; \backslash times\; \backslash frac\{1\}\{3n^2\}$
" $\backslash times\; HH$, 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 $\backslash frac\{1\}\{4\}$ of the content of a cone similar to the vessel, or an altitude equal to the height of the frustum. In a sphere, it is $\backslash frac\{1\}\{4\}$ 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 subtracting 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 subtracted 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 .079,577, and the product by the length. Ex. Let the mean circumference of a tree be 10 3 feet, and the length 24 feet. Then $10\backslash frac\{1\}\{3\}\; \backslash times\; 10\backslash frac\{1\}\{3\}\; \backslash times\; .079,577\; \backslash times\; 24\; =\; 202.615$, 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 .0795,774,715, 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 cubical 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$\backslash frac\{1\}\{2\}$ grains. The Scots Troy-pound (which, by the statute 1718, was to be the same with the French) is commonly supposed equal to 15$\backslash frac\{1\}\{2\}$ ounces English Troy, or 7560 grains. By a mean of standards kept by the Dean of Guild at Edinburgh, it is 7599$\backslash frac\{1\}\{2\}$ 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$\backslash frac\{1\}\{2\}$ pounds.”

GEORGE,