Εικόνες σελίδας
PDF

peridicular to the line of section, DE. Make the ordinates qi, rk, si, &c. each respectively equal to the ordinates Qa, Rb, Sc, &c; and through the points D, i, k, I, &c, to E, draw a curve, which will evidently be the section of the cylinder, as required.

The same may be done in this manner, viz.—Bisect the line of section DE in the point /. Draw tm perpendicular to DE. Make tm equal to the radius of the circle which forms the end of the cylinder; then, with the axis major, DE, and the semi-axis minor, tm, describe a semi-ellipse, which will be the section of the cylinder required.

A DEFINITION.

221. A Cuneoid is a solid ending in a straight line, in which, if any point be taken, a perpendicular from that point may be made to coincide with the surface: the end of the cuneoid may be of any form whatever.

The cuneoid, which occurs in architecture, has a semi-circular or a semielliptical end, parallel to the straight line to which the perpendicular is applied.

Problem 39.

222. To find the section of a cuneoid, with a semi-circular base, the given data being a section through the axis, perpendicular to the vertex, or sharp end, and the line of section upon that end.

Let ABC, (Ar. 6, pi. W,) be the section through the axis, perpendicular to the sharp edge, and let DE be the line of section.

This construction is similar to that of finding the section of a cylinder, excepting that, instead of drawing parallel lines from the base, AB, they are, in this figure, drawn from the points of section in AB to the point C, which is the vertex of the cuneoid: the ordinates, Qa, RJ, SC, &c, being transferred, respectively, to qi, rk, si, &c; and the curve D, i, k, l, &c, to E, being drawn through the points, D, i, k, l, &c, by hand.

Problem- 40.

223. Given the position of the seats of three points, in the circumference of the base of a cylinder, and the lengths of the perpendiculars intercepted between the points and their seats, to find the section of the cylinder passing through these three points.

Through the three points, A, B, C, {fig. 7, pl. VI,) describe the circumference of a circle. Join the two remote points, A and B, and draw AD, CF, and BE, perpendicular to AB. Make AD equal to the height upon A, BE equal to the height upon B, and CF equal to the height upon C. Produce BA and ED to meet each other in H: draw CG parallel to BH, and FG parallel to EH. Join GH. In GH take any point, G, and draw GK perpendicular to CG, cutting BH in K: from the point K draw KI, perpendicular to EH, cutting EH in L. From H, with the radius HG, describe an arc, cutting KI at I. Join HI. In the circumference, ACB, take any number of points a, b, c, &c, at pleasure, and draw ae, bf, eg, &c, parallel to GH, cutting AB at e,/, g, &c Through the points e,f, g, &c, draw lines ei,fk, gl, &c, parallel to GK, or AD, or BE, cutting DE at i, k, l, &c; from the points of section, i, k, l, &c, draw the lines in, ko, ip, &c, parallel to HI. Transfer the ordinates, ea, fb, gc, &c, to in, ko, m, &c; then, through the points D, n, o, p, &c draw the curve Dnqp, &c, to E, and it will be the section cut by the plane, as required.

Problem 41.

224. Given the great circle of a sphere, and the line of position of a section at right angles to that great circle, to find the form of the section.

Let ABC, {fig. 8, pl. VI,) be the great circle, and AB the line of section. On AB, as a diameter, describe a semi-circle, which will be the section required: since all the sections of a sphere, or globe, are circles.

Problem 42.

225. Given the section of an ellipsoid, passing through the fixed axis, and the line of position of another section, at right angles to the first section, to find the form of the section through that line.

Let ABCD, [fig. 9, pl. W,) be the section through the fixed axis, and EF the line of position. Through the centre of the ellipsoid draw AC parallel to EF. Bisect EF in H, and draw HG perpendicular to EF. Find HG a fourth proportional to AC, DB, HE. Then, with the axis major, EF, and the semi-axis minor, HG, describe a semi-ellipse, and it will be the section of the ellipsoid required.

If AC be the axis major, BD will be the axis minor. In this case, join DC, and draw EG parallel to DC; then HG will be the height found geometrically.

Problem M.

226. To find the section of a cylindric ring, perpendicular to the plane passing through the axis of the ring, the line of section being given.

Let ABED, {fig. 10, pl. VI,) be the section of the ring, passing through its axis, and let AB be a straight line, passing or tending to the centre of the two concentric circles, AD and BE; also, let DE be the line of section. On AB describe a semi-circle, and take a, b, c, &c, any number of points in its circumference; draw the ordinates, ae, bf, eg, &c Through the points, e, f, g, &c, in the diameter AB, draw the concentric circles, ei, fk, gl, &c, cutting the sectional line DE in the points i, k, l, &c Through the points, i, k, l, &c, draw in, ko, Ip, &c, perpendicular to DE; transfer the ordinates ea,fb, gc, &c, of the semi-circle, to in, ho, m, &c: and, through the points D, n, o,p, &c. draw the curve, DnopqE, which is the section required.

PLANE TRIGONOMETRY.

DEFINITIONS OF TERMS IN TRIGONOMETRY.*

227. The Complement Of An Arc is the difference between that arc and a quadrant or quarter of a circle.

Thus, the arc BC, which is the difference between AC and AB, is the complement of AB; and AB is, in like manner, the complement of BC.

* Trigonometry is that branch of Geometry which treats exclusively on the properties, relations, and measurement, of triangles.

[graphic]
[graphic]
[graphic]
[ocr errors]

228. The Supplement Op An Arc is the remainder between that arc and a semi-circle.

Thus, the arc given being AB, its supplement is BC.

229. The Sine Of An Arc is a straight line, drawn from one extremity of the arc, upon and perpendicular to a radius or diameter.

Thus, BM is the sine of the arc AB; and here it is evident that an arc and its supplement have the same sine.

230. The Cosine Of An Arc is the sine of the complement of 0' that are Hence, BO or IM is the co-sine of the arc AB; and, therefore, the sine of the complement BC

231. The Tangent Of An Arc is a straight line, drawn from one extremity of the arc, where it touches it, to meet the prolongation of the radius through the other extremity.

The line AK, touching the arc at A, and extended to meet the radius IB produced, is the tangent of the arc AB.

232. The Co-tangent Of An Arc is the tangent of c the complement of that arc

Thus, CL is the cotangent of the arc AB, or the tan gent of the arc BC.

In the annexed diagram let AB, AC, AD, AE, AF, AG, AH, to A, be the several portions of the circumference, by supposing the point B to revolve round the circumference from A to B, C, D, E, F, G, H, the sine of any arc, in the first quadrant, increases from A to C, where it is the greatest possible, and then decreases to E, where it becomes zero; the sine will, therefore, be positive for the first semi-circumference, and in the other half it will be negative.

[graphic]
[graphic]

The cosine will be positive in the first quarter, negative in the second and third, and again positive in the fourth.

The tangent will be positive in the first quarter, negative in the second, positive in' the third, and negative in the fourth.

TRIGONOMETRY.—THEOREM 1.

233. If a perpendicular be drawn from an angle of a triangle, to the opposite side, which is the base; then, as the base is to the sum of the two sides, so is the difference of the sides to the difference of the segments of the base.

For, {theorem 62, page 56) AC2-CD*=AD'

and, again, {theorem 62) BC*- CD2=BD2.

Subtract the second equation from the first, and the

result is AC*-BC2=AD*-BD2:

but, since the difference of the squares of any two quantities is equal to a rectangle contained by their sum and difference;

therefore (AC+BC)(AC -BC) = (AD+BD)(AD-BD)

Whence, (theorem 40, page 41) AD+BD : AC + BC :: AC - BC : AD - BD.

[graphic]

TRIGONOMETRY.—THEOREM 2.

234. The sum of the two sides of a triangle is to their difference as the tangent of half the sum of the angles at the base is to the tangent of half their difference.

Let ABC be a triangle; then, of the two sides, CA and CB, let CB be the greater. Produce CA to E, and make CE = CB, and join BE. Produce BC to F; and, through A, draw FD, perpendicular to EB, meeting it in D; then FBD will be half the a^sum of the angles at the base, and ABD half their difference. Likewise, DF is the tangent of the angle FBD, and AD the tangent of the angle ABD: moreover BF is the sum of the two sides BC, CA, and AE is their difference.

Then, by similar triangles, BFD, EAD, BF x AD = FD x EA. Wherefore BF : AE :: FD : AD; which is the proposition to be demonstrated.

[graphic]
« ΠροηγούμενηΣυνέχεια »