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tion of the helix. And any number of consecutive revolutions may be drawn by setting off the constant distance a'a" along the projections of the generatrixes, from each point of the first, which is obviously equivalent to a repetition of the construction by which the first convolution was obtained.

Let it now be required to determine the tangents to the helix parallel to any given plane, aßa', the horizontal trace of which may be assumed perpendicular to xy; but the construction will be equally simple if the plane have any proposed position.

Draw the tangent bd to the circle abc, and make it equal to the arc ab rectified. The vertical plane passing through bd would be tangential to the cylinder in the generatrix 6 6'6", and the helix would develop on this plane into the tangent to the curve at 6,6', which would pass through the point d,d'. b'd', therefore, drawn through d', the vertical projection of d, will be that of the tangent sought, and bd will be its horizontal projection.

A line parallel to this tangent through the point 0,6', would have e for its horizontal trace, and if this line were rotated round the axis of the cylinder, it would generate a right cone, having a circular base of the radius oe, the generatrixes of which cone are, as has been stated, parallel to the tangents of the helix: consequently the plane b'fh, which passes through the vertex of this cone, and which is parallel to the given plane a' ßa, must cut the cone in lines parallel to the required tangents; these lines have og, oh, for their horizontal projections.

And since the tangents to the helix have tangents to the circle abc, for their horizontal projections, if the tangents mr, ns, be drawn respectively parallel to og, oh, they will be the horizontal projections of the tangents sought. The vertical projections m', n' of the proposed points being determined by means of the horizontal ones, m'r', n' s' are to be drawn parallel to Ba or bf, for the vertical projections of the tangents to the helix.

Although two other tangents to the circle abc might be drawn, parallel to og, oh, respectively, yet they would not be the horizontal projections of two tangents to the helix parallel to the generatrixes of the cone; it being obvious, from the nature of the curve, that the tangents to it at those points would be inclined in the opposite direction to that of the corresponding generatrix.

There would, however, be two tangents parallel to the two first at each successive spire of the helix, at points having m for their common horizontal projections.

PROBLEM 6.

To construct the projection of a spherical epicycloid, and to

determine a tangent to it at any proposed point.

56. Def. If a circle roll on the circumference of another, which remains stationary, so that the two circumferences are always tangential to each other, any point in that of the first will describe a curve termed an epicycloid : if the moving circles always remain in the same plane with the fixed one, it is clear that the epicycloid will be a plane curve lying in the same common plane. But if the two circles are not in the same plane, but constitute the bases of two right cones having a common vertex, and therefore equal generatrixes, and the one cone roll over the surface of the other fixed one, so that they may always have the same common vertex, and touch in a line common to both, then the epicycloid is termed a spherical one.

For, on this supposition, it is clear that the fixed circle is on a sphere, having the common vertex of the cones for its centre, and their common generatrix for its radius; and the moving circle always remains on this spherical surface, or, in every position, lies entirely in it.

[graphic]

The plane of this moving circle, in every position of it, cuts that of the fixed base in a line which is a tangent to both circles, and it is obvious that the inclination of the two planes remains constant, and equal to the angle formed by the axes of the two cones.

The tangent at any point of the spherical epicycloid must be in the tangent plane to the sphere, and it will now be proved that it also lies in a plane tangential to a second sphere, which would have for its centre the point in which the moving circle touches the fixed one in that position.

Instead of circles, let two polygons * ABC.., A'B'd'.., be supposed to roll one on the other; and to simplify the subject, let the polygons be supposed regular ones of any number of sides at pleasure, the sides of the one being equal to those of the other, although the number of them may vary in the two, and let the angle a' coincide with a; the polygon ABC being assumed as fixed, let A'B'd turn round the point A, so that b' may coincide with b: then again the polygon turns on b, till d' coincides with c, and so on in succession.

Now, any point m of the moving polygon, during the rotation of it on the angle a, will lie on the surface of a sphere having a for its centre, and am for its radius; and when the point o becomes the centre on which the polygon turns, m will lie on the surface of a sphere having b for its centre and BM for its radius, and so on in succession. The tangent, consequently, at any point of the line generated by the point m, ought to lie in the tangent plane to a sphere, the centre of which is the angular point of the polygon, about which the rotation is effected that brings the point m into its actual position. This conclusion being independent of the magnitude of the sides of the polygon, must apply equally in the case of two curves rolling on each other, that is, the moving curve being in any, whatever, of its positions, and if h express the point in which it touches the fixed curve, and n the corresponding situation of the gene rating point m: the tangent at the point n of the curve generated by the point m, will always be contained in the tangent plane at the point n to the sphere having u for its centre and un for its radius.

. To draw the projection of the Epicycloid. This projection may be made on the plane of the base of the

* To avoid unnecessary complexity, the polygons are not drawn in the figure.

fixed cone; let the circle abb, a, described on the horizontal plane with the radius oa, be the base of this cone, and let the point which generates the epicycloid have originally been at a, and the point of contact of the two circles be supposed now at b; to determine the actual position of the generating point a second plane of projection must be assumed cutting the first in ob, it must consequently pass through the common vertex 0,0' of the two cones, and will cut the moving cone in the isosceles triangle o'bc', which is one of the data of the problem ; the horizontal trace of the plane of the base of this cone will be the line bc, a tangent to the fixed circle in the point b, while bc is obviously the vertical trace, and at the same time the vertical projection of the base of the cone.

The circle bmc, the diameter bo of which is equal to that of the base of the moving cone bc', will be that base turned down on the horizontal plane. Make the arc bm of this circle equal to the arc ba, for since the point m was originally at a, and the different elements of the two arcs bm and ba, having been successively applied to each other, these two arcs must be equal.

Make bm' in bc' equal to the distance of m from bc; through mi draw a line parallel, and through m one perpendicular to bc; these two lines will cut each other in my the projection of the point m on the horizontal plane, and therefore a point in the projection of the epicycloid.

By a repetition of this construction, as many points in the projection as may be deemed necessary may be determined, but instead of repeating the construction at different points of the circle abb'a', as for example at b, all that part of the construction which requires the circle bmc, may be made on that circle ; thus, the arc bm is to be made equal to ba, and the horizontal projection m, of the point

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