It is evident that a system of jointed bars having the shape of the funicular polygon would be in equilibrium under the action of the given forces, supposed applied to the joints; moreover any bar in which the stress is of the nature of a tension (as distinguished from a thrust) might be replaced by a string. This is the origin of the names "link-polygon" and "funicular" (cf. § 2). If funiculars be drawn for two positions O, O´ of the pole in the force-diagram, their corresponding sides will intersect on a straight line parallel to OO´. This is essentially a theorem of projective geometry, but the following statical proof is interesting. Let AB (fig. 27) be any side of the force-polygon, and construct the corresponding portions of the two diagrams, first with O and then with O´ as pole. The force corresponding to AB may be replaced by the two components marked x, y; and a force corresponding to BA may be represented by the two components marked x´, y´. Hence the forces x, y, x´, y´ are in equilibrium. Now x, x´ have a resultant through H, represented in magnitude and direction by OO´, whilst y, y´ have a resultant through K represented in magnitude and direction by O´O. Hence HK must be parallel to OO´. This theorem enables us, when one funicular has been drawn, to construct any other without further reference to the force-diagram. [Illustration: FIG. 27.] The complete figures obtained by drawing first the force-diagrams of a system of forces in equilibrium with two distinct poles O, O´, and secondly the corresponding funiculars, have various interesting relations. In the first place, each of these figures may be conceived as an orthogonal projection of a closed plane-faced polyhedron. As regards the former figure this is evident at once; viz. the polyhedron consists of two pyramids with vertices represented by O, O´, and a common base whose perimeter is represented by the force-polygon (only one of these is shown in fig. 28). As regards the funicular diagram, let LM be the line on which the pairs of corresponding sides of the two polygons meet, and through it draw any two planes [omega], [omega]´. Through the vertices A, B, C, ... and A´, B´, C´, ... of the two funiculars draw normals to the plane of the diagram, to meet [omega] and [omega]´ respectively. The points thus obtained are evidently the vertices of a polyhedron with plane faces. [Illustration: FIG. 28.] [Illustration: FIG. 29.]