24 
AC 3= log d x at A — log d^ at C 
== whole pressure 
x ord. = whole pressure 
t of uniform atin. 
: 30 in. : height 
30 : 27863 ft. 
= 64020 
.43429 
AC = 64020 (log d-log d) 
l a 
== 64020 (log h — log K) 
1 \ 
The following is an abstract of the Paper : — 
Let R T represent the surface of the Earth ; AB, BC, CD, &c, thin 
strata of air, extending to Z, of equal thickness. Let d ± be the density or 
weight of the small column, AB ; d^ of BC, &c. Then it can be shown 
that d. : d n : : d • d • : d Q id , &c. Therefore the densities are in geome- 
±2 2 o o 4j 
trical progression decreasing, while AB, AC, AD, &c, are in arithmetical 
progression increasing. Or, beginning at K, both series are increasing. 
Of these two series, the arithmetical is the logarithm of the geometrical. 
If we had corresponding tables, we could know AC by subtracting log c? 3 
at C from log. d at A, i.e. 
AC = log. ^ at A — log. d 3 at C. 
Now let the densities Kk, Hh, Gg, &c, be drawn so that KH, KG, 
&c, shall be proportional to the logarithms of Kk, Hh, Gg, &c, and 
draw a curve through k, h, g, &c, then the curve, which is a hyperbola, 
will be the atmospheric logarithmic curve. 
Next — the density at every point of AK = ordinate of that point, 
. * . sum of densities, which equals the weight or pressure at A = area on Aa 
between the curve and the perpendicular. The same at C — Area on Cc ; 
hence these areas represent the pressures respectively. 
If a tangent at a be drawn, meeting the asymptote in m, Am is the 
sub-tangent, and Cn is the sub-tangent of c ; all sub-tangents are equal. 
