458 



CRYSTALLOGRAPHY. 



M L'hsma- 



tlCdl 



Theory. 



Plate 

 CCXXtll. 



Fitf. 35. 



3b". 



Pig. 33. 



breadth by the same number of ranges on all the angles 

 of the parallelopiped (Fig. "5.) ; and let us take, as an 

 example, that which takes place upon the angle BCD. 

 Let C kl be the measuring triangle, in which C k mea- 

 sures the distance between the point G and the first 

 plate of superposition ; kl is applied to the lateral face 

 of that plate, and measures its height, and C / coincides 

 with the secondary face, produced by the decrement 

 under consideration. 



Having drawn the diagonals db, fh (Fig. 36.) on 

 the bases of the molecule, let fall cl perpendicular to 

 db, and xz perpendicular both to d/i and to fh Let 

 N be the number of ranges abstracted. We will have 

 Ck (Fig. 35.) = N X ct (Fig. 36.), and kl (Fig. 35.) 

 r= xz (Fig. S6. ) ; and the angle C kl (Fig. 35.) will be 

 equal to that which the plane bdfli (Fig 36.) forms 

 with the plane fgh. But as these three quantities are 

 known, hence it will be easy to find kCl (Fig. 35.) 

 which measures the inclination of the new face to the 

 parallelogram ABCD. The effect of the decrements on 

 the other angles is calculated in the same manner. 



Let us now consider the hypothesis in which the 

 decrements that take place upon the two angles DCG, 

 BCG have such a ratio to that which takes place upon 

 the angle BCD, that the faces produced by the three 

 decrements coincide in one plane. Let AG (Fig- 38.) 

 be the nucleus as before. Let us suppose the decre- 

 ment which takes place in breadth upon the angle 

 BCD has such a measure, that the edge of the first plate 

 of superposition passes by m r. In which case, each of 

 the lines Cm, Cr will include as many lengths of mo- 

 lecules equal to cd or cb, (Fig. 36.) as there are ranges 

 abstracted by the decrement. Having taken upon CG 

 (Fig. 38.) a part Cc equal to eg, (Fig. 36'.) let a plane 

 pass through the points m, c, r. This plane is parallel 

 to the face that will be produced by decrement. To 

 prove this, draw the indefinite lines ms and ru parallel 

 to CG, and prolong them above the nucleus, so as to 

 make Mot or Rr equal to Cc. It is evident that Mm 

 and Rr represent two faces situated on the side of the 

 first plate ; therefore the face produced by the decre- 

 ment passes through the points M, R. It likewise pas- 

 ses through C, which is the point from which the de- 

 crement set out ; therefore the plane MCR coincides 

 with the face produced by the decrement. But the 

 lines Cc, Mm, Rr, being three longitudinal edges of the 

 molecules, situated parallel to each other between the 

 two planes mcr, MCR, these two planes must be paral- 

 lel. That is to say, mcr is parallel to the face produ- 

 ced by the decrement. 



The same reasoning applies to the hypothesis when 

 the decrements take place in height. In that case, it 

 would be necessary, in order that the plane mcr were 

 parallel to the face produced, to have cm=cd, (Fig. 36.) 

 er=cb, and that the line Cc (Fig. 38.) should contain as 

 many times eg (Fig. 36.) as there are ranges subtracted 

 in height. 



Suppose the plane MCR to be prolonged above the 

 faces CDFG, BCGH, and let us consider these pro- 

 longations as two faces that would have the effect of two 

 decrements, the one upon the angle DCG, the other 

 upon BCG. These decrements being equal, we may 

 confine ourselves to that which takes place upon the 

 angle DCG. Since the plane cmr is parallel to the 

 face which results from the decrement, it is clear, that 

 c m coincides with the inferior border of the first plate 

 of superposition applied upon CDFG; and that Cr 

 contains as many lengths of molecules as there are mo- 

 Imiles subtracted in height* 



If the decrement relative to the angle BCD takes Matliema- 

 place by one range, it is evident, that the two other de- 

 crements on the angles DCG and BCG will likewise 

 take place by one range, because, in that case, the three 

 lines Cm, Cr, Cc, being each equal to the length of 

 one molecule, must of course have the same measure. 



But if the decrement relative to the angle BCD take 

 place by more than one range, then the two odiers will 

 be necessarily intermediate, and it will be sufficient to 

 have the law of the first decrement to determine the 

 two others. Suppose, for example, that the decrement 

 on the angle BCD takes place by three ranges in breadth. 

 In that case, Cm and Cr will be equal each to the 

 length of three molecules, and Cc will be equal to one 

 length. Then the decrement on the angle DCG takes 

 place in such a manner, that there are three lengths of 

 a molecule subtracted along CD, and only one along 

 CG ; and this decrement takes place by three ranges in 

 height, since Cr corresponds with three lengths of a 

 molecule. The same rule holds with the decrement on 

 the angle BCG. 



In all cases of this nature, the theory considers only 

 the decrements which take place according to the ordi- 

 nary laws ; because a much more simple solution re- 

 sults from it. The two other decrements are consider- 

 ed as subsidiary, and as coming to second the effect of 

 the first, and to continue it over the parts adjacent to 

 the face which it has produced. 



The greatest number of faces which a secondary 

 crystal can have on the hypothesis of a decrement, on 

 all its angles, is 24 ; because there are eight solid angles, 

 composed each of three plane angles, which are the 

 terms from which as many decrements set out. The 

 smallest number of faces, according to the same hypo- 

 thesis, is eight. And though, in fact, there are always 

 24 decrements, yet eight only are considered ; which 

 enables us to apply the ordinary laws to determine the 

 figure of the secondary crystal. The most simple case 

 is that in which the nucleus is a cube, and all the de- 

 crements take place by one range. The result is the 

 regular octahedron, as we saw in the last Chapter. 



But it may happen that the three decrements which 

 take place round tlie same solid angle, are intermediate. 

 In that case, it is sufficient to determine one of them, 

 in order to be able to judge of the two others, by means 

 of a construction similar to that which we have already 

 employed. 



Let Fig. 3. represent the nucleus, marked with let- "Plm ?. 

 ters, according to the rules laid down in the last Chap- IVXXlv". 

 ter for writing the symbols of crystals. Let us conceive *>' S- 

 that a decrement takes place on the angle O, ascending, 

 which produces a face parallel to the plane n rs, and of 



2 



which the expi-ession is (O D 3 F 4 ). From this it fol- 

 lows, that On = 3cd, (Fig. 36.) Or=4>cb, and 0*= P " TE 

 2 ' ' CCXXIIfc 



The expression of the decrement on the left of the '^' * 

 angle O will be ( 4 D 3 H 2 ) and tliat of the decrement 

 on the right of the same angle (O 3 F 4 FF )'. 



To determine the angles which the faces produced 

 by intermediate decrements make with the correspond- 

 ing faces of the nucleus, the most simple method is to 

 consider each little group of molecules which results 

 from the decrement, as forming a single molecule, which 

 brings the calculus to that which is employed for the 

 ordinary decrements on the angles. Let us take, as an 

 example, the decrement upon the angle O, ascending, 



represented by the symbol (O D 3 F 4 ). It is easy to 

 judge, that in this- case, the group which represents the 



