the Strength atkl Values of Spirituous Liquors, 331- 



latter '7965. Taking proportional parts, as before, '7972 

 in Column VIII answers to 887 in Column II. 



Therefore, bv the iirst method, l6iJ-42 x '7972, or 

 129'4S, is the per-centage required. 



And, by the second method, to log. •7973 = 1*901 5673 

 Add the constant log. = 2-210630S 



And 129'48, the natural number answering" 



to the sum, is the pcr-centage required V2* 1121981 

 = 29-48 O.P. 



•ed V2- 



2. What is the pcr-centage of a liquor whose specific 

 gravity at 75° is 953i? 



Solutioft. 



On looking into the Columns II and VIII of the tabic 

 marked " hkat 75°", we find that the number in the latter 

 corresponding to 953'^ in the former is -3827. 



Therefore, by the first method, 162-42 x '3827, or 

 62*16, is the required pcr-centage. 



And by tlie latter, log. -3827 = 1*5828585 

 Constant loo-. = 2*2106309 



And the same number 62*16 answering to^ 



the sum, indicates the per-centage as be- > 1*7934893 

 fore (= 37-84 U.P.) J 



§ 33. Problem TIL — The per-centage of a liquor at a 

 given temperature being given, to Jind its specific gravity 

 at that temperature. 



Practical P»ules. 



Multiply the per-centage by -006157; search for the 

 product iu Column nil of the table adapted to the given 

 temperature, and the correspondent number in Column II is 

 its specific gravity. 



Or, add the constant logarithm 3*7893692 to that of the 

 per-centage; search for the correspondent natural number 

 in Column Fill of the proper table, and the specific gravity 

 against it is that of the liquor. 



Examples. 

 I. What is the specific gravity of a liquor at 48°, which 

 is 129*48 (or 29-48 O.P.) at that temperature? 



Solution, 



