28 THE MECHANICS OF THE EARTH's ATMOSPHERE. 



From the preceding it results that the pressure of the air against a 

 plane disk turned normally towards it is 



_ 2.2G4 + 00942 x p , 

 ~ 1,000,000 '^^' 



Where D is expressed in old Prussian loths and |;, F, and c in [Rhenish] 

 inches. According to the above, the pressure against a square disk of 

 1 square foot area, moving with a velocity of 50 feet per second, would 

 for example be 140.8 loths, or nearly 4.4 pfund. 



For reduction to metric measures and weights I take not the metre 

 itself but the decimetre as the unit of measure for lengths and surfaces, 

 in order to remain within the limits of the observations. Therefore 

 the resistance of the air for a temperature of 15^ G. and a barometric 

 pressure of 28 Paris inches,* expressed in grammes, amounts to 



(0.00707 + 0.0001125i>) Fc\ 



Where p represents the circumference of the disk, F the sectional area, 

 and c the velocity expressed in decimetres. 



The pressure that very small disks experience when struck normally 

 by a current of air is also given by another simple consideration, whose 

 correctness has in general been confirmed by many experiments. These 

 experiments indeed are limited, so far as known, to streams of water; 

 but the expansibility of the air is certainly in this case without intiu- 

 ence, since the observations mentioned in the preface, upon the direc- 

 tion and strength of currents deviated in front of opposing disks, 

 showed identical results with water and with air. 



Imagine a vessel filled to the height h with a fluid of which one unit 

 of volume or 1 cubic inch weighs ;/ loths. The bottom of the vessel 

 therefore experiences on each square inch a pressure equal to ;//<, wheik 

 no side pressure exists. If there is suddenly made therein an oi)ening 

 of 1 square inch, the outflow of the fluid through it begins with the 

 velocity c = 2\/.(//it, and if we catch the stream hy an equally large sur- 

 face directed normally against it, then the pressure D upon this is again 

 equally as great as before upon the bottom of the vessel, namely, yh. 

 From this we have 



D=yh = l^c 



For the density of the air above adopted its specific weight is 

 0.001223; therefore a cubic inch weighs 0.001495 loth, and g is equal to 

 187.6, if the semi-acceleration due to gravity is expressed in inches. 

 From this we have these results : 



D = 0.000001992 = 1.992 millionths of a loth. 



* The density is that of air at 15° C. and 28 Paris inches or TST.Qe™™ under gravity 

 at Berlin (52° 30 ), but strictly speaiiing the pressure should be stated in standard 

 measure as 758.47™°' under gravity at 45'^ and sea level. 



t g is the height fallen through in 1 second, or one-half the acceleration due to gravity. 



