. for the Pressure of highly compressed Steam, Gases, fyc. 39 



taken into account as well as the pressure of the air in the 

 tube C ; otherwise it will not be a correct indication of the 

 force of the gas or fluids acting upon the surface of the mer- 

 cury in the chamber B. Therefore, in the following calcula- 

 tion for graduating the scale attached to the tube C, a proper 

 allowance is made for the weight of the column of mercury. 



Annexed is the investigation of a theorem for graduating 

 the scale. The table was computed therefrom, and a scale 

 agreeable thereto is made to be attached to an instrument 

 which I am now making, for the purpose I have described. 



Calculation for the Scale. 



Ascertain nicely the capacity of the chamber A, together 

 with the tube C and its ball D, and let that quantity be de- 

 noted by M. Then ascertain the capacity of the tube and 

 ball only, and denote that by m : and let the ratio of these two 



quantities be denoted bj r r; that is r= — . 



And let a denote, in inches, the whole length of the tube C, 

 including a length of tube equal in capacity to the ball D ; 

 and let x represent the height, in inches, of the mercury in 

 the tube C, from the bottom of the tube at a. Then when 

 the mercury just reaches the bottom of the tube at a, it is 

 plain that the pressure must be equal to r atmospheres. And 

 when the height of the mercury in the tube is equal x, then 



the pressure on the inclosed air will be equal — ^— x r at- 

 mospheres. 



But in order to ascertain exactly the force of the gas acting 

 upon the surface of the mercury in the chamber B, we must 

 add to the above the weight of the column of mercury above 

 the level in the said chamber B. Let b denote the number of 

 inches that a, the bottom of the tube, is above the level of the 

 mercury ; and let c represent the height of a column of mer- 

 cury equal in weight to one atmosphere, say 29*5 inches. 

 Then the weight of the column of mercury in the tube C will 



be = — - — . Therefore putting y equal the required number 

 of atmospheres, we shall have 



a x + b ^. 



X ?• -I = y. Or, 



*= — 7 2 — ± *v — - -f car + ab — acy. 



(No. 2) 



Now put r the ratio = 20- and r/ = 76'-8, the whole length of 

 tin- IuIm- including the ball at top; and put c = 29'5 = the 



height 



