BY A SPACE ROTATING ABOUT A PLANE. 1 1 



of the side-cubes of the eightcell by the rotating space. We will 

 close this first section by explaining this. 



We consider that side-cube of the eightcell of which the lozenge 

 ABCD (fig. 2) is a section, one of the three side-cubes projecting 

 themselves on the rectangle PP 2 Q 2 Q (fig. 1). The space of that 

 cube meets the plane % in the line H X H 2 (fig. 2); this side of 

 the hexagon Ii Q joins the centres H x , H 2 of the two adjacent faces 

 of the cube projecting themselves on to the rectangle P l P 2 Q 2 Q 1 

 (fig. 1) the centre O of which forms the projection of tt and there- 

 fore of H X H 2 . 



We project the chosen side-cube on to the plane passing through 

 PQ and the centre of the cube — and containing therefore the 

 opposite edge which again may be called P 2 Q 2 (fig- 5) — as a 

 rectangle PP 2 Q 2 Q with the sides PQ = 3 J/2 cm., PP 2 = G cm. and 

 remark that the centre O of P 1 P 2 Q 2 Q 1 — where P x and Q t 

 bisect PP 2 and QQ 2 — is still the projection of H x and H 2 , 

 and therefore of H x H a . This projection PP 2 Q 2 Q (PP 2 = 6 cm.) differs 

 from the projection PP 2 Q 2 Q (PP 2 = 2 \/6cm.) of fig. 1 and 

 fig. 3 , which difference is due to the fact that the planes of projection 

 differ. Indeed the plane of projection t' of fig. 1 and fig. 3 passes 

 through PQ and the centre of the eightcell, whilst that of fig. o 

 is determined by PQ and the centre of the chosen side-cube; the 

 first plane t' indicates the angles of rotation of the intersecting 

 space turning about t, the second cannot do this. So the question 

 arises: „how can we find the lines of intersection l x , l 2 , ... / 7 ' 

 of the new plane of projection with the seven considered positions 

 of the rotating space?" To obtain an answer we remark that any 

 line of this new plane of projection, e.g. the diagonal PP 2 of the 

 face of the cube which projects itself on PP 2 , is divided by the 

 different points of intersection with the lines l x ', l 2 , ... / 7 ' we 

 are bent on determining, in the same proportion as the projection 

 PP 2 of that selfsame diagonal on %' is divided by the correspon- 

 ding lines l t , L, ... / 7 of fig. 3. After having obtained these 

 lines (fig. 5) the construction of the side-faces is very easy. Indeed 

 we know that the normal erected in the threedimensional space of 

 the cube on the plane PP 2 Q 2 Q in any point £7 within that rectangle, 

 reckoned from U to either of the points of intersection with the 

 boundary of the cube, is equal to the distance of that point U 

 from the nearer of the two edges PQ, P 2 Q 2 - So in the case III 

 the half C 3 D 3 E 3 W 3 of the pentagon B. à C 3 D 3 E à F 3 of fig. 3 is 

 obtained by erecting in M 3 and W 3 normals M.^D^ = P X P 

 and W^~E % = W^P on / 3 ' ; so are found the halves A x C x D x , C 2 D 2 P , 



