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and recourse must then be had to the normals as above described.


To determine the intersection of two surfaces of revolution,

the axes of which lie in different planes.

54. The horizontal plane must be again assumed perpendicular to the axis of the first surface, and the vertical plane parallel to the axes of both; the data will present the same arrangement as in the preceding problem ; only the horizontal projection of the axis of the second surface will not pass through the point a, but will be parallel to xy, as bc.

Since the axes lie in different planes, there can be no two parallels, one on each surface, which lie in one plane, or are comprised in one sphere; another proceeding must therefore be adopted,—that of cutting the two surfaces by horizontal planes. Each of these will intersect the first in a circle, the horizontal projection of which will be an equal circle, which is readily described; but the section of the second surface will be a curve, the projection of which must be obtained by means of points, through which this curve is drawn; the points in which it cuts the circular section of the first surface will be points in the projection of the common insersection sought; and the vertical projections of these points will lie in the trace of the auxiliary plane. Every two points thus to be determined will necessitate the construction of a curve.

Let d'f', parallel to xy, be the vertical trace of an auxiliary horizontal plane; its intersection with the first surface is projected vertically in dd", and horizontally in the circle dmm, described from a as a centre, with įd'd" for a radius ; the vertical projection of the curve of the section with the second surface will be f'f"; and to obtain points in the horizontal projection, the surface must be supposed cut by planes perpendicular to its axis, forming consequently circular sections.

Let g'g" be the vertical projection of such a circular section, the point h', in which f'f" meets gg", is the vertical projection of two points in the auxiliary curve required ; consequently the horizontal projections of these two points must lie in h'h, drawn perpendicular to xy. Let the plane of the circle gʻg", be supposed turned round on the line h' h till it is horizontal, and then projected on the horizontal plane ; it will be seen that this projection must be a circle, described from the point k, with kl for its radius. Now, the points projected in k', before mentioned, are not moved on this supposition; their horizontal projections, therefore, will be hh, in which the circle kl cuts h'h.

By means of other planes perpendicular to b'c', other points of the section corresponding to f'f" are determined; and the curve itself hfh obtained. This curve and the circle dmm, are then the horizontal projections of the sections of the two given surfaces made by the auxiliary plane d'f'; and hence m, m', m', m", in which these lines cut each other, must be two points of the intersection of those surfaces.

These constructions must be repeated for other auxiliary planes, till a sufficient number of points are determined *.

The tangent to the intersection of the two surfaces is obtained, as before, by means of normals. The vertical projections of the parallels of each surface in which the point of contact is situated, must first be determined, and subsequently those of the normals; but instead of finding the traces of the plane of the normals, which would be an unnecessary operation, the intersections of this plane with the horizontal and vertical planes, passing through the point in which the normal to the second surface meets its axis, are determined.

PROBLEM 5. To construct the projection of a helix on a vertical cylinder,

and to determine the tangents to this helix, which are parallel to a given plane.

55. Any whatever cylinder being given, let it be supposed cut by a plane perpendicular to its generatrixes; if distances

* To avoid complicating the figure, the constructions for finding the points, m, m', m',m", are alone shown, and not the intersection of the two & Irfaces.

are set off along these generatrixes from the plane proportional to the arcs of the base comprised between these generatrixes, and any fixed point of this base, assumed at

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pleasure, the curve which would be thus determined is a helix*.

* If a right-angled triangle, the base of which is equal to the circular base of a right cylinder rectified, and its height any whatever, be conceived as wound round the cylinder, so as to coincide everywhere with its surface, the hypotheneuse will form a helix on the cylinder. The side of the triangle will coincide with the directrix passing through the point of departure of the curve.

This mode of generation shows that if the cylinder be developed on a plane, the base becomes a right line, perpendicular to the generatrixes; and the lielix, consequently, develops into a straight line, oblique to these generatrixes. Consequently all the tangents to the curve are inclined to the generatrixes in the same angle. If, therefore, we develop the cylinder and the helix into a plane, touching the cylindrical surface in the generatrix, passing through the point of contact, the helix when developed will coincide with the tangent to the curve at that point.

Another property directly resulting from these conditions is, that if all the tangents to the helix were moved parallel to their original position, but so as to pass through one common point, they would form a right conical surface, the axis of which would be parallel to the generatrixes of the cylinder. Hence it follows that the tangents parallel to a given plane must be parallel to the right lines, which would be formed by the section of this cone by a plane parallel to the given one.

Let the base of the cylinder be the circle abc lying in the horizontal plane, and let a be the origin of the arcs or circular abscissa. And let a'a" be the distance between any two consecutive revolutions of the helix*. The circumference abca, and the altitude a'a", must be divided into any number of equal parts, twelve for example. Draw the vertical projections of the generatrixes passing through the points of division of the base, and make the heights of these projections in succession equal to the corresponding segments of a' a", by drawing lines parallel to xy through the points of division of that line. The line a''c'a" drawn through these points will be the vertical projection of one convolu

* This distance a'a" will therefore be the altitude of the rightangled triangle alluded to in the preceding note..

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