1825.] Barometrical Measurement of Heights, 83 



of mercury to be equal to the mean density of the equiponderant 

 column of air (a supposition which will not at ail affect the 

 result), if we divide the two columns, whose heights will now be 

 equal, into the same number of exceedingly small strata, weigh- 

 ing alike, then will some one or other of the strata of the air, 

 varying in depth, be sensibly of the same thickness, or vertical 

 height, as any one of the uniformly deep strata of the mercury. 

 The density of this stratum is evidently equal to the mean den- 

 sity of the column of air, and khowing the number of the strata 

 it is distant from the summit of the column of mercury, we 

 easily ascertain the pressure it sustains by adding the sum of 

 the equal weights of these strata to the height of the barometer 

 at the upper station. With the assistance of a table of loga- 

 rithms, together with its modulus^ the mathematician would 

 readily determine that the pressure supported by a volume of air 

 uniformly of a density equal to the mean density of a stratum of 

 dry air intercepted by the pressures of 15'5 inches and 30*5 

 inches would be 22*1602 inches, and that the height of the 

 object, or of the column of air at 32° F. would be equal to 30*5 



minus 15*5= 15 inches of mercury multiplied by ^ of 12,000 



inches, or to 17,666 feet. Had the pressure expressive of the 

 mean density been 26-0988 inches, the length of the column of 

 air would have been 12,000 times 15 inches, or 15,000 feet; but 

 as the heights are inversely as the pressures, its altitude must 

 be increased in the ratio of 22-1602 to 26-0988. 



But we may calculate the altitude of the object by a process 

 more generally intelligible, and which will have the advantage 

 of demonstrating that the pressures or heights of a barometer 

 carried successively to a series of stations uniformly increasing 

 in their distance from the surface of the earth, will decrease in 

 geometrical progression. Our plan will be to divide the column 

 o^ mercury into a sufficient number of equal parts or heights, to 

 which we must affix the corresponding intercepting pressures. 

 The columns not exceeding one inch in height, the mean pres- 

 sure, or half the sum of the pressures at the base and the sum- 

 mit, will not materially differ from (exceed) the pressure supported 

 by a stratum of air of which the density, supposed to be uniform, 

 is equal to the mean density of the column of air counterpoising 

 that inch of mercury. As the altitude of a stratum of dry air 

 under the pressure of 26*0988 inches counterpoising an inch of 

 mercury is equal to 1000 feet, and as the heights are inversely 

 as the pressures, we ascertain the altitude in ieet corresponding 

 to the different columns of mercury by dividing the constant 

 number 26098*8* by the respective mean pressures. The sum 

 of these altitudes will be equal to the altitude of the object. 



* For air containing its mean quantity of moisture, and to include some corrections 

 for gravity, substitute 26210. 



g2 



