( ^510 ) 



8. Finally we consider a tangential net. 



Two right lines through P which intersect at right angles are 

 touched by one conic of the net. If the right angle included by 

 these lines turns about the vertex P the pairs of tangents drawn 

 through Q to the variable conic form an involution. For every ray 

 s, drawn through Q, determines a tangential pencil belonging to the 

 net, having the pairs of an involution (r, r-') in common with the point 

 P. The conic touched by the orthogonal pair (r^ r') has a second 

 tangent s' through Q, forming with s a pair of the involution indi- 

 cated above. The orthogonal pair (s, s') determining a conic for 

 which r and r' are at right angles to each other, we can draw but 

 one orthoptical circle through the points P and Q. 



According to the circles (ro) possessing an orthogonal circle or 

 intersecting a fixed circle in two diametrically opposite points or 

 passing through a fixed point, the obtained net of circles is repre- 

 sented by an orthogonal hyperboloid of revolution with one or two 

 sheets or by an orthogonal right cone. 



Chemistry. — "Oh solubility and meltingpoint as criteria for 

 distinguishing racemic combinations^ fseudoracemic mixed crystals 

 and inactive conglomerates." By Prof. Bakhuis Roozkboom. 



Though several times attention has been drawu to the phenomena 

 of solubility and melting in order to distinguish between the types 

 mentioned above, no certainty has been attained as yet. 



1. Solubility. "We only get a clear insight in the phenomena of 

 solubility by drawing attention to the number of solubility curves 

 obtainable at a given temperature. 



If (Ja is the proportion of the 

 saturated solution of the dextrosub- 

 stance D, and Ob the same for the 

 laevosubstance L, these two are equal 

 at the same temperature. 



By adding L to the Z)-solution 



and vice-versa we now get, if no 



racemic combination appears at the 



temperature used, nothing else than 



two solubilitycurves, starting from 



^ a the points a and b and meeting in 



Fig. 1. c. Their precise direction depends on 



the manifold actions that can take place in the solution. From 



the perfect equivalence of L and D it results however necessarily 



that c must always lie on the line OB, which halves the angle of 



