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Problem, IIL 



The height at which mercury ftands in two barometers, one 

 at the fummit of any elevation, and the other vertically under 

 it, or at leaft not more than ten miles diftant from a vertical 

 fituation (the nearer the better), and the mean temperature being 

 32° ; to find the meafure of that elevation, and the weight of 

 ©f the intermediate column of air. 



Solution. 



Find the logarithms of thofe mercurial heights. The difFer- 

 ence of the four firft figures on the left of thofe logarithms 

 gives the meafure of that elevation in Englifli fathoms j and the 

 difference of the remaining three figures gives the decimal parts 

 of thofe fathoms, the whole multiplied into fix gives feet, or 

 multiplied into 72 gives inches, as already fa,id. The height of 

 the mercury in both barometers fliould be taken at the fame 

 inftant, after allowing them fufficient time to aflume the tem- 

 perature of the air, or at leaft to cool. 



To find the weight of the intermediate column of air, ob- 

 ferve by how many inches, or parts of an inch, the mercury 



