On Crystallography. 129 



method of indicative signs. Let us conceive that there is 

 made upon the angle O, ascending, a decrement which pro- 

 duces a face parallel to the plane n r s, and the expression 



of winch is (OD*F 4 ), whence it follows that on = 3crf 

 {fig. 2), Or = 4cb, and Os= Qcg. 

 This being done, the expression of the decrement on the 



left of the angle O will be (^OD J H 2 ), and that of the de- 

 crement to the right of the same angle (0 T F 4 H>). 



25. Tn order to determine the angles formed by the faces 

 produced by the intermediary decrements with the corre- 

 sponding faces of the nucleus, what presents itself as the 

 simplest is to consider every little group of molecules, 

 which results from the decrement, as forming one single 

 molecule ; which brings back the calculation to that which 

 is employed for the ordinary decrements on the angles. 



Let us take for example the decrement on the angle O 

 i 

 ascending, represented (OD*F 4 ). Tt is easy to judge that 

 in this case, the group which represents two subtractive 

 molecules placed the one above the other, is that which we 

 sec fig. 6, and in which the side mn is composed of three 

 ridges of molecule, the side np of four ridges, and the 

 side nk of two ridges, on account of the decrement by 

 two ranges in height. 



Having traced on the bases the diagonals mp 9 i o, I 

 draw vt perpendicular upon wp, then us perpendicular as 

 well upon mp as upon io. 



Let n ty (fig. 7) be the mensurator triangle, in which 

 nt being regarded as lying on the plane AEOE (fig. 5) 

 will be equal to the same line (fig. 6). Besides, we shall 

 have ty (fig. 7) = u s (fig. 6), and the angle nty (fig. 7) 

 will be equal to that formed by the plane mpoi (fig. 6) 

 with the triangle i k o. Thus it will be easy to find the 

 angle yn t (fig. 7) which measures the inclination wanted. 



26*. The solutions of problems of this kind are often sim- 

 plified in practice, by a series of the regular form of mole- 

 cules. Let us suppose, for example, that the latter are 

 cubes. Let us designate each of their ridges by unity. 

 We shall have (fig. 6) m n = 3, np = 4, nk = 2, mp = 



V (mn) 1 + hip) 1 — V 25 = 5. nt — = ---.«$ = 72 A = 2. 



x ' v r/ rnp 5 



Thus also nt (fig. 7) = -7-, ty = 2. 

 Vol. 36. No. 148. August 1810. I Thus 



