CHAPTER III

THE PROJECTIVE PROPERTIES OF RECIPROCAL FIGURES

49. The Genesis of Reciprocal Figures.-Let the arrowheaded lines 1 and 2 (Fig. 68) represent the lines of action and directions of two given forces. Then if, by the wellknown theorem of the triangle of forces, we draw the lines AB and BC (Fig. 69) parallel to the directions of those forces and of lengths respectively proportionate to their magnitudes, the closing line AC will represent both the direction and magnitude of the resultant. The line of action of this resultant will traverse the point i of intersection of the components 1 and 2, and be parallel to AC.

In order to construct what is termed the funicular polygon of the system, pitch a pole at any point O (Fig. 69), and from O draw rays to the three corners ABC of the polygon of forces, which rays will be hereafter described by the symbols O31, 012, 023; then in Fig. 68 draw the line 3'1' parallel to O1, and intersecting the lines of action of 3 and 1 in the points 3' and 1' respectively. From l' draw 1'2' parallel to O1, meeting the line of action 2 in 2'. Lastly, from 2' draw a line 2'3' parallel to O meeting the line 3'1', first drawn, in 3' and completing the funicular 3'1'2'.

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50. The point of intersection (3) of the closing lines of the

funicular polygon lies upon the line of action of the resultant. It is evident that the line of action of the resultant must pass through i; and, further, that the direction of this

resultant must be parallel to the line AC. It only remains to shew that the straight line joining i and 3' is parallel to AC.

By way of premise it is necessary to prove that the sixsided figure, formed by joining in pairs any four fixed points in a plane, is determined by any five of its sides; or, in concrete terms, the six-sided figure formed by lines joining in pairs the four fixed points 3'1'2'i (Fig. 68) is fully defined before the sixth or last line connecting the points 3' and i has been filled in.

THEOREM I.-Given, then, five sides of the complete quadrilateral a'd'b'e' (Fig. 70) parallel respectively to the five sides of

the similarly lettered quadrilateral adbe, to prove the sixth side b'd' parallel to the sixth side bd. From the given similarity of the figures eabce and e'a'b'c'e' and the admitted parallelism of lines ad and a'd' in all positions, it follows that the rays ad and a'd' revolve at uniform rates about their respective centres a and a'; or, in other terms, describe equal angles in equal times and, therefore, cut off similar intercepts in equal times upon the given parallel lines ec and e'c'.

Wherefore (Euclid VI. 2), if the lines bd and b'd' are parallel in any one position, they will be parallel in all positions. But in the limits, when bd coincides with be or be and b'd' with b'e' or b'c', b'd' is given parallel to bd; therefore b'd' must always be parallel to bd.

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Now, in order to apply this theorem to the case in hand, draw oa (Fig. 68) parallel to 1'3' or to Os (Fig. 69); and let this line oa meet the diagonal bi produced in a. Similarly draw be parallel to oi or BC, meeting the other diagonal in c. Then, by the theorem just proved, since the five lines joining the four points O, A, B, C are parallel to the five lines joining the similarly lettered points oabc, the sixth line ac must be parallel to the sixth line AC. It still remains to be proved that i3' is parallel to ac, for which purpose it is necessary to establish the following proposition.

THEOREM II.-If the opposite sides (oi and b3') of a complete quadrilateral (oib3') be interchanged; that is, if oa be drawn parallel to b3' meeting the diagonal bi in a; and if bc be drawn parallel to oi meeting the second diagonal in c, the sixth side ac of the second quadrilateral so formed will be parallel to the sixth side i3' of the first or original quadrilateral. For, if x be the point where ab and oc intersect, the triangles x3′b and xoa are similar; so that

Again, since the triangles xoi and xcb are similar, we have

xi xb

хо xc

whence, by multiplication of ratios,

xi x3'

- ; ха хс

(2),

wherefore (Euclid VI. 2) the lines ac and 3'i are parallel.

But it has already been shewn that ac is parallel to AC; hence 3'i is parallel to AC.

From this investigation it follows that, if five sides of Fig. 68 be parallel to five sides of Fig. 69, the sixth side 3'i must be parallel to the sixth side AC. But five sides of Fig. 68 have been drawn parallel to five sides of Fig. 69; viz.

wherefore line 13' must be parallel to AC.*

Q. E. D.

By combining any fourth force with the resultant 3, and treating the new pair 3 and 4 in precisely the same way as the old pair 1 and 2, it may be shewn that the principle holds for any number of forces. Hence we may conclude that the line AC joining the loose ends of the polygon of any number of forces furnishes the magnitude and direction of the resultant of those forces; whilst the point of intersection 3′ of the closing lines 1'3′ and 2′3′ determines a point on the line of action of the resultant. When, therefore, a system of forces is in equilibrium,

1o. The force and funicular polygons both close,

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2o. The rays О31, 012, O2 furnish the stresses set up in the sides 3'1', 1'2', and 2'3' of the funicular, according to the same scale that the lines 123 (Fig. 69) represent the forces applied in the directions 123 (Fig. 68).

51. Definitions of Reciprocity.—The companion figures 68 and 69 are said to be reciprocal in various senses. Perhaps the most practical definition is to term them reciprocal in the two senses, (1°) that their corresponding lines are all either parallel or perpendicular to each other, (2°) that all lines forming a pencil of rays in one figure form a closed polygon in the other. But Culmann, taking a

*Theorem I. of this original demonstration was published by the author in the Engineer of 14th September 1888; Theorem II. is now published for the first time.