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is contained in the last of the integral parts of the first arc, with again some remainder; find in like manner how often this last remainder is contained in the former; and so on continually till the remainder becomes too small to be taken and applied as a measure. By this means we obtain a series of quotients, or fractional parts, one of another, which being properly reduced into one fraction, give the ratio of the first arc to the semicircle, or of the proposed angle to two right angles, or 180°, and consequently that angle itself nearly in degrees and minutes.

Thus, suppose the angle BAC be proposed to be measured. Produce BA out towards f; and from the centre A describe the semicircle abcf, in which ab is the measure of the proposed angle. Take ab in the compasses, and apply it four times on the semicircle as at b, c, d, and e; then take the remainder fe, and apply it back upon ed, which is but once, viz. at g; again take the remainder gd, and apply it five times on ge, as at h, i, k, l, and m; lastly, take the remainder me, and it is contained just two times in ml. 4, 1, 5, 2; consequently the fourth or last arc

1 5

A

a B

Hence the series of quotients is em is the third ml or gd, and

therefore the third arc gd is or of the second arc ef; therefore again this

second arc ef is

is

1

413

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or of the first arc ab; and consequently this first arc ab

or of the whole semicircle af. But of 180° are 37 degrees, or

37° 8' nearly, which therefore is the measure of the angle sought. When the operation is carefully performed, this angle may be obtained within two or three minutes of the truth.

In fact, the series of fractions forms a continued fraction. Thus, in the example above, the continued fraction, and its reduction, will be as follows:

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1 13 = ; 4+1+5/2 4+1 411 63

the quotients being the successive denominators, and 1 always for each nu

merator.

Scholium.

This method is due to De Lagny, and a corresponding process has been applied by Adams to finding what portion any given line is of another given line. The problems are so precisely alike, that any specific detail is altogether unnecessary. Mr. Sankey has also applied the metallic cycloid to the measurement of circular arcs: but of course its determinations are only rough approximations, though their being obtained with great facility, and near enough for most practical purposes, is a recommendation to its familiar usage.

PROBLEM LI.

To find the diameter of any solid sphere, as a ball or shell.

FROM any point P on the surface of the given sphere, and with any convenient radius (about the estimated chord of 60° will generally be best for accuracy) describe a circle ABC, and in it take any three points (nearly equidistant by estimation) A, B, C; and on

paper describe a triangle abc whose sides are equal to those of ABC: about abc describe a circle abcd, and draw its diameter bd: with centres b, d, and radii equal to the linear distance of P from A, B, C, describe arcs intersecting in ƒ: and lastly, about bfd describe the circle bfd. Its diameter will be equal to that of the given sphere.

For let PO be drawn perpendicular to the plane of the circle ABC meeting it in O: then, being produced to Q, it will be a diameter of the sphere. If also BD be a diameter of the circle ABC, it is equal to the circle described about abc, since circles described about equal triangles are equal.

Again, if PD, PB, be joined, the three sides of the triangle DPB are equal to the three sides of the triangle bfd, each to each: and hence the circles about them are equal, and their diameters also equal. But PQ is the diameter of the sphere, and hence of the circle DPQ, the plane of which passes through PQ: and hence, again, the diameter of the circle bfd which is equal to that of the circle DPQ is also equal to the diameter of the given sphere.

Scholium.

This problem, which has often been proposed as a new one, owing to its not being inserted in books which are generally consulted, is yet as old, at least, as the first century B.C.: for it is found in the Spherics of Theodosius, and constructed almost exactly in the same way as above.

PRACTICAL GEOMETRY IN THE FIELD.

THE absence of instruments in cases of exigence renders it of much importance to be possessed of means of determining approximately the positions of certain inaccessible points with regard to others that are accessible, by having recourse only to lineal measurements made in accessible places. A few of the more useful problems of this nature are annexed; but the want of sufficient space prevents the insertion of a greater variety.

All the solutions here given are effected by means of staves set up at particular and specified stations, together with the use of the chain or other lineal measure; as the determinations are practically made with greater certainty by this than by any of the means usually employed. So far, however, as other modes are concerned, there are solutions of the main part of these problems, though not specified in reference to this use, to be found amongst the problems in Practical Geometry already given.

PROBLEM I.

To continue a straight line on the ground, the two determining points, A, B, of which are given, there being no visual obstacles intervening.

Fix upright staves* at A and B, and walk as nearly as you can judge in the required prolongation to any point C'; which we will suppose to be a little to the right or left of the actual prolongation of AB. This will be known

ΧΗ

B

* It is important in all these problems that the staves be placed as nearly perfectly vertical to the horizon as possible.

by the visual lines C'A, C'B, not coinciding. If the left side of the staves be visible, we are then on the left of the line AB, and if the right, we are on the right of AB. Move slowly towards the line, till the visual lines from A and B coincide, as at C; then C will be in the prolongation of AB. If we wish to place a mark as at H between two points A, B, in lineation, we must first lineate to C beyond one of them; and then by C, B, lineate H by a subsequent, and similar, process.

Scholium.

It will often in the following problems be necessary to find the intersection of two lineations; and though the process is very simple, it may be well to explain it in such a way as to be effected with the least possible trouble.

B

EH

CASE I. When the lineations AB, CD, do not intersect within the figure ABCD, bounded by lines joining each of the stations to the adjacent ones. Let the point H be taken in lineation of AB beyond the probable point of intersection of AB, CD; then if the observer walk from H towards B in lineation, till the points C, D, also appear in lineation, to E, the point E will be that required.

G

CASE II. When the intersection G of AC, BD, are required. Set up two auxiliary marks, K and F, in lineation of AC and BD respectively; then by means of B and F let the observer walk in lineation of DB, till he arrives in lineation of AC, as indicated by the marks C, K. The point G, where this occurs is the intersection of AC, BD.

Many obvious facilities may be brought into the operation when there is more than one observer; but these solutions are adapted to the most unfavourable case as to assistance.

PROBLEM II.

To find the lineation of two points B, C, when obstacles intervene G, which render the points B, C, invisible.

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First method. Take any point A without the line BC, from which B and C can be seen; also points c and b in them; find the intersection D of Bb and Cc, and then the intersection of cb and AD. Calculate the distance bd from the formula obtained below, and measure that distance in lineation of bc: the point d will be in lineation with BC. Produce AD to meet BC in a: then (th. 97) we have cE: Eb:: cd: db, or

CE

B

D

a'a

Eb.bc cE - Eb

C d

- Eb Eb cd - db (= cb): bd. Hence bd = ; and all the lines on the right side of the equation are measurable, and hence bd can be computed.

Scholium.

Should a second point be required for the purpose of continuing the lineation, we may repeat the process with another, or with the same triangle, ABC. The latter will be the better method, and is thus performed :

Retaining one of the marks c, change the other from b to b': then find the

B

D

point D', the point E', and finally the distance b'd' from the above formula with the new values of the several parts of the line cb'. Second method, without any calculation. Take any point A, and in the lineations AB, AC, take any points c and b: find the intersection D of the lineations Bb, Cc; take any point G in AB, and find the intersection E of Gb, AD, and the intersection F, of cE, AC: then the intersection H of bc, FG is in lineation with B and C.

C

E

For, produce AD to meet BC in a: then (th. 97) the lines cb and GF divide BC in prolongation in the same ratio that BC is divided in a.

Hence they cut

BC in the same point, and each other in lineation with B and C*.

PROBLEM III.

Through a given point B to lineate towards the invisible intersection H of two given lineations FG, bc. (Fig. prob. 2, second method.)

TAKE any point A and mark F, b, in any lineation through A, and the points G, c, in AB: find the intersection E of Gb, Fc, and D the intersection of AE, Bb: then cD, Ab intersect in a point C, which is in the same line with the given point B and the invisible point H.

This depends on the same principle as the last, and is proved in the same way.

PROBLEM IV.

To find the length of a line AP, inaccessible at one extremity.

First method, when one end of the line is accessible. Take any convenient station B on the ground in lineation with A, P, and a station R out of that line; prolong BR to any convenient point C: then marking the point Q where the lines RP, AC, intersect, we shall find AP by the equation

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For, (th. 95) PA: PB:: QA.RC: QC.RB; or div°. PA PB - PA :: QA.RC: QC.RB QA.RC, whence the theorem follows.

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Second Method. Take any two points, B, C, lineating with A, and any point F in AP: find the intersection E of BF, CP; the intersection D of BP, CF; and the intersection G of DE, AP: and measure GF, FA. Then

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* On the same principle we may resolve the following problem, which at first sight may appear altogether impossible.

The directions of two lines whose point of concourse is invisible, and the directions of two others in like circumstances, are given: to find a point in lineation of the two invisible points of concourse Let Ac, bD, be two lines whose point of concourse B is invisible, and cD, Ab, two others whose concourse, C, is also invisible: to find a point d in lineation of the invisible points B, C. Produce the four lines backwards till they form the quadrilateral AcdB; draw the diagonals meeting in E, and find bd by the equation of the text: then d is in lineation of B, C.

This problem is one of frequent occurrence in practical lineation, on account of visual obstacles occurring to prevent B and C from being seen from any point in lineation with them.

For, (th. 95) PA: PG :: AF: FG, or PA- PG: PA :: AF – FG : AF,

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One or other of these methods, where the relative positions of the fundamental points, from their entirely arbitrary character, admit of indefinite variation, will apply to all cases whatever.

PROBLEM V.

Through an accessible point Q to draw a line parallel to the accessible line AB. TAKE any three convenient equidistant points A, P, B, in the given line AB; take any point C in the lineation AQ, and find the intersection D of CP, BQ; and then the intersection R of BC, AD: the lineation QR will be parallel to the lineation AB.

P

R

For, (th. 96) AQ: QC:: AP.BR: PB.RC, and AP = PB (by constr.): hence AQ: QC:: BR: RC, and (th. 82) the line QR is parallel to AB.

When the line AB is inaccessible, the solution will be effected by prob. 9.

PROBLEM VI.

Through a given point A to lineate parallel to two given parallel lines

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For by a converse course of reasoning to that employed in the last problem, it may be shown that CDd would bisect ab and QR; and by similar reasoning to the last, that AB is parallel to either QR or ab, and hence to both of them.

Scholia.

This process is often applicable under circumstances where the last cannot be applied, arising from the intervention of practical obstacles. For instance, the lines might not be traceable so as to get the parallel through A to the line ab; whilst it might be possible (by the preceding problem) to trace a parallel to ab through Q, and thence by this to also trace a parallel to ab. This latter part, too, where two parallels already exist, is attended with some practical convenience, of which the preceding is destitute; as it requires not even the use of the chain and if Q, R, b, are remarkable points, we may even solve the problem if they be inaccessible.

Two parallel lineations may be traced upon the ground by the following considerations:

1. The lineations of a very distant object will be sensibly parallel.

2. The shadows of two upright staves taken at the same time are parallel; as are likewise the lineations of any star taken at the same time.

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