CRYSTALLOGRAPHY. 



459 



Matkertia. 



tfcal 



Theory. 



Fig. g. 



molecule subtracted, is that which is represented in 

 Fig. 4, in which the side mn is composed of three 

 lengths of molecules, the side np of four lengths, and 

 the side nk of two lengths, in consequence of a decre- 

 ment by two ranges in height. 



Having drawn upon the bases the diagonals mp, io, 

 let fall nt perpendicular to mp, and us perpendicular 

 both to mp and io. Let nty (Fig. 5.) be the measuring 

 triangle, in which nt being conceived to coincide with 

 the plane, AEOI (Fig. 3.) will be equal to nt. (Fig. 4.) 

 We will likewise have ty (Fig. 5.) = us, (Fig. 4.) and 

 the angle nty (Fig. 5.) will be equal to that which the 

 plane mpoi (Fig. 4.) makes with the triangle iho. 

 Hence it will be easy to find the angle ynt (Fig. 5.) 

 which measures the inclination wanted. 



The solution of problems of this kind is often sim- 

 plified in practice, in consequence of the regular form of 

 the molecules. Let us suppose, for example, that they are 

 cubes ; and let us give to each of their edges the value 

 of unity. We have, in such a case^ (Fig. 4.) ?nn=:3, 



M»=4, nk=2, mpz= V {uny -\- [n py — V25z=5 ; n/= 

 mn%np 



= V 1 ; and u s =r n Jc = 2. Hence, n t ( Fig. 5. ) 



= y and ty =. 2. And 



nt : ty : : y : 2 : : 6 : 5. 

 In this case the angle nty is right. Hence we see how 

 easy it is in such a case to find the angle ynt. 



The measuring triangles, relative to the decrements 

 on the angles, may be substituted for those which we 

 have considered in the decrements on the edges, and 

 will serve equally to determine the secondary forms. 

 Suppose, for example, that AG ( Fig. 6. ) represents a 

 cubic nucleus, which undergoes decrements by two 

 ranges upon the four edges of the base ABCD, and that 

 we wish to know the angle of the pyramid jADCB 

 produced by this decrement. Having drawn the dia- 

 gonals BD, AC, let fall from their point of intersection 

 o the line op perpendicular to CD. Draw sp. Upon 

 op take the part pr, equal to two lengths of a molecule, 

 and from r draw ru perpendicular upon the plane 

 ABCD, and which, by hypothesis, is equal to the 

 length of one molecule. The triangle upr will perform 

 the function of the ordinary measuring triangle ; and 

 by means of the right angle urp and the ratio 2 : 1 be- 

 tween the sides pr and ur, it will be easy to find the 

 incidence of DsC on the base ABCD, as well as the 

 values of the other angles. For on account of the simi- 

 lar triangles up r, spo, the whole is reduced to calculate 

 the angles of a right pyramid, in which the side BC of 

 ihe base, which is double of po, is to the axis os as 4 

 to 1. 



On the other hand, if we take upon Co the part Cm 

 equal to two diagonals of the molecule, and from the 

 point n raise n z perpendicular upon ABCD, C n will 

 represent the distance of the point C from the first plate 

 of superposition, taken in the direction Co, and n z will 

 be equal to the length of a molecule ; hence it follows, 

 that the triangle zCn may also perform the office of a 

 measuring triangle. We will have Cra : nz : : 2 V2~ : 1 ; 

 and because the triangle zCn is similar to the triangle 

 sCo, the question considered in this new point of view 

 is reduced to find the angles of a right pyramid, in 

 which the_demidiagonal C o of the base is to the axis 

 o s as 2s/ '1 : 1. It is usual in this manner to substitute 

 one measuring triangle for another, when there results, 

 in consequence, a greater facility of calculation. 



The preceding details may be considered merely as 



introductory observations, in order the better to explain 

 the nature and importance of the measuring triangles. 

 Let us now proceed to the method of calculating, and 

 let us begin with the rhomboid, which (including the 

 cube,) affords the easiest application of the theory, and 

 at the same time the most varied results. 



The first object to be attained, is the algebraic ex- 

 pressions for the principal lines in the rhomboid. Let 

 Fig. 7- represent an obtuse rhomboid, because it occurs 

 most usually; though the algebraic expressions will ap- 

 ply equally to any rhomboid whatever. Draw the di- 

 agonals bfad. The horizontal demidiagonal be or 

 c/'is called g, while the oblique demidiagonal ac or ad 

 is called p. 



Let adsg (Fig. 8) be a four-sided figure formed 

 by two opposite oblique diagonals ad, gs of the rhom- 

 boid, (Fig. 7-) and by the edges a g, d s included be- 

 tween these two diagonals. Such a figure is called the 

 principal section of the rhomboid. 



From the point d draw d r perpendicular to the axis 

 a s, and from the point g, drawg n, likewise perpendi- 

 cular to the axis, and continue it till it meets ad. This 

 line will bisect ad. For if the diagonals fg and b.g 

 (Fig. 7.) be drawn, the whole line g c (which is the 

 same as in Fig. 8.) will be situated in the plane bfg 

 (Fig. 7.) which passes through the point c. It is evi- 

 dent, likewise, that c n, which is a perpendicular drawn 

 from the centre of the equilateral triangle b fg, is half 

 the length of g n, which goes from the centre to one of 

 the angles of this triangle. Now, gn is called the per- 

 pendicular to the axis, and c n the semiperpendicular to 

 the axis. 



The following are the algebraical expressions for the 

 different edges of the rhomboid, in functions of g and 

 P- 



1. ab (Fig. 7.) = 'V / (frc)' + c ) : = v V +P t - 



2. bf=2g; cg=V'$g r ; c»=s v'-j-g 2 . 



3. The perpendiculars gn and dr (Fig, 8.) divide the 

 axis into three equal parts. For the similar triangles 

 ac n, adr, give a d : a c : : a r : a n. But a dz=2 a c, 

 therefore a rz=2 a n. And the triangles dsr, gan be- 

 ing similar and equal, we have r s=a n. Therefore, 

 an=?i r=rs. But a n=V / («c)'- — (e n)-=</p' ! — 4g 3 . 

 Hence, a s—S'/p* — }g°-~ */g p l — \ g z s= 'fgjfl — 3g ! , 



Problem. Given the demidiagonals g and p to de- 

 termine, in a general marner, three species of angles, 

 namely, the plane angles of the rhomboid, the incli- 

 nation of the respective faces, and the angles of the 

 principal section. 



1 . For the plane angles, draw a m (Fig. 7. ) perpen- 

 dicular to df, am will be the sine of the angle afd, 

 supposing af as radius. Let us find the ratio between 

 af and the cosine fm. 



We have already af=z\/g' t -\-p i . 

 bfxac U ? 2 p \ 



d J g- + F 



Hence, fm or v/(«/y_ («,„)* = z l> /g--+p'-—.tilPl 



_ / g<-2g*p* + p* 



V g' + f ' 



Hence it follows, that af-.fm : : g 2 +p* : =i=g*rz=M* 

 The upper signs of the last term belonging to the case 

 when the rhomboid is obtuse, the lower to that in winch 

 it is acute. 



This result gives us a remarkable property of the 



Mathema- 

 tical 

 Theory. 



Alfrebraic 

 expression* 

 for the 

 rhomboid. 



Pl.ATZ 



CCXXIV. 

 Fig. 1. 



Fi?.& 



