Ueber den russiscuen Zirkon. 



Für die doppelte Neigung der Fläche der (i riindpyramide о = P zur 

 Verticalaxe. 



Am Krystall Nr. 1. 



0, : 03 





95° 



42' 



0" (П 



(Оі):(оз) 





95 



40 



10 







95 



40 



0 



Mittel 



= 



95° 



40' 



5^(2) 



0-2 : 0,, 





95° 44' 10" 







95 



45 



0 



Mittel 





95° 



44' 35\з, 



(02) : Ы 



= 



95 



42 



0 







95 



38 



40 



Mittel 





95° 



40' 20" 



0| : (03) 



_ 



84 



20 



50 







84 



20 



0 



Mittel 





84° 



20' 25^' also 



Complément 





95 



39 



35 



(0,) : oo 





84 



19 



55 







84 



20 



0 



Mittel 





84° 



19' 



58;' also 



Complément 





95 



40 



2 (6) 







84 



21 



0, also 



Complément 





95 



39 



0(7) 



04 : (02) 





84 



14 



40, also 



Complément 





95 



45 



20 



02 : ilfg 





132 



10 



0, also 



0 : 

 an derS 

 Spitze) 

 03 : Л/3 





95 

 132 



40 

 11 



0 (9) 



30, also 



an (1er) 

 Spitzej 





95 



37 



0 (10) 



Am Krystall 

 0, : (03) = 84° 

 Complément = 95 

 0,: = 132 



an der> 

 Spitzej 



Nr. 2. 



21' 20;' also 



38 40 



11 0, also 



38 0 (,2\ 



(0,) : Ж, 



an der) 

 Spilzej 



132'' 

 95 



Am Krystall Nr. 

 95° 41' 30" 



30; also 



0 M3) 



3. 



mit â Fernr. 





Am Krystall Nr. 4. 



0, 



03 = 95° 46' 0'\,5, 



Ы 



(03) = 95 38 0 (,e) 



0-2 



oj, = 95 40 30 





K) = 95 40 0 



0-2 



(0,) = 84 20 0, also 



Complément = 95 40 0 ^,9, 





(02) = 84 21 45, also 



Complément = 95 38 15 (20) 



Am Krystall Nr. 5. 



Ol 



03 = 



95° 



38' 40" (2,) 



(Ol) 



(Оз) = 



95 



40 0 ,22) 



02 



04 = 



95 



44 0 (23) 





К) = 



95 



44 45 (2.) 



03 



{o^) = 



84 



24 0, also 



Complément = 



95 



36 0 J25) 



02 



Ы = 



84 



20 0, also 



Complément = 



95 



40 0 ,26) 



0, 



ы = 



84 



14 20, also 



Complément = 



95 



45 40 (27) 



0, : 





132 



8 30, also 



an der > z 

 Spitzej 



0, : M, 



: 95 

 : 132 



an der) = 95 

 Spitzej 



(0,): M, = 132 



an der) 

 Spitzej 



43 

 7 

 44 

 12 

 95 36 



, also 



0 



40, i 



40 (29, 



0, also 



0 (30) 



