THE SLIDE RULE IN ROCK ANALYSES 565 



index of 'X" is under ''E" and then set ^'E" on 552 on "D." In 

 this position it crosses "C" at 19.98. 



Total on "C" 



Separate Percentages 



icite — 36. 19 per cent 



K2O — 7 . 78 per cent 





AlA-8.43 " " 





SiO— 19.98 " " 



Total — 36. 19 per cent 



The reverse operation, or conversion of minerals to their con- 

 stituent oxides, will now be described. Suppose we take the case 

 just cited. A rock contains 36.19 per cent leucite. What are 

 the corresponding values of K2O, AI2O3, and Si02 ? The percentage 

 composition of leucite is 



K2O— 21.5; Al A— 23-3; SiO^— 55.2 



Set "E" over an index on ''D," and since there is a choice of two 

 such positions (one at each end of the scale) suppose the right index 

 of "D" is chosen. Then move the slide so that the leucite per- 

 centage, 36.19 on "C," is under "E." It will be noted that of 

 the three oxide values — 21.5, 23.3, 55.2 — only that of silica can 

 be found on "D" under the slide in its present position. Hence 

 set "E" on 552 on "D" and it will be found to cross "C" at 19.98, 

 the percentage value of SiOa. Then shift the slide so that 36 . 19 on 

 "C" is opposite the left index on "D." Set "E" on 215 on ''D" 

 and it will cross "C" at 7.78, which is the potash value in the 

 leucite percentage. Move "E" to 233 on ''D" and it will cross 

 "C" at 8.43, which is the AI2O3 value desired. 



This explanation covers any ordinary case, including both the 

 synthetical and the analytical processes. If the following point is 

 borne in mind there should be no confusion. In the method as 

 outlined, all percentage values from the table were read on "D," 

 whereas all analytical values and their equivalents were read on "C." 

 The opposite way is equally feasible, but if one way is always 

 followed there is little excuse for confusion. Similarly, all these 

 operations can be performed on scales "A" and "B," but the divi- 

 sions are smaller and not so easily read. 



