The contact of the side C A of the triangle with the vertex S is continuous, along with that of the other two sides with the arcs which form the sides of the duangle; the angle between each pair of consecutive normals is therefore always 120, because they remain always perpendicular to the sides of the triangle. Thus the conditions necessary to the continuous restraint of sliding are fulfilled by this pair of figures, and consequently (by 21) the normals must always intersect in one point, so that the figures will serve as the basis qf one of the higher pairs of elements. We have now to find the corresponding centroids.
Fig. 95.
Fig. 96.
(a.) Centroid of the Triangle. In order to make our investi- gation as general as possible we shall take the problem in the form used above, of angle and triangle, considering the line P Q as a plane figure (see 5), the motion of which relatively to the figure for which the angle U TQ stands is the same as that of the duangle to the triangle. We require to know, as was shown in 8, the paths of at least two points in the moving figure. We do know, how- ever, the rectilinear paths P T and Q T of the points P and Q. The normals to these paths cut each other in 0, Fig. 95, and this point must lie in the circle already found, for the angle P Q enclosed by the normals is obviously equal to 180 a. Further, both OPT and Q T being right angles, the line joining and T must be the diameter of the circle P T S Q. The chord P Q and the angle a being constant, the size of this circle is fixed, ,the distance TO of the instantaneous centre from the point T is constant; the centroid is therefore a circle described about T
