TOO SMITHSONIAN MISCELLANEOUS COLLECTIONS VOL. 51 



be the horizontal velocity of flow and e' the density of the air at the 

 altitude h above the bottom of the valley, therefore, for the point 

 whose abscissa = — A/ 4; then will the total quantity condensed 

 per second above the base area one meter broad from the beginning 

 of the clouds to the point x, expressed in grams, be as follows: 



H 



G s = {e'u'g x (h)dh (20) 



The quantity of air, e u kilograms, flows in one second through a 

 strip of the vertical plane at x = —A/ 4, having a unit width and the 

 height d h; but an equal quantity must flow out per second through 

 the vertical whose abscissa is x, and since the condition is steady, 

 it therefore behaves as though the quantity of air, e u, had moved 

 in one second along the line of flow from — A/ 4 up to x; but in this 

 the quantity of water e u g x (h) is separated from the air according 

 to our definition of g. 



If we have computed G as a function of x, according to formula 

 (20), then, finally, we have 



W = dG (21) 



dx 



as the quantity of water, expressed in grams, per horizontal square 

 meter per second, that falls at the place x. In this way the deter- 

 mination of Wis executed more conveniently than through the 

 direct formula (17). By assuming the average conditions for the 

 summer in the above example for a = 1, we find that the integral 

 (20), if we compute it as approximately equal to the sum of the 

 intervals between the individual current curves of flow as con- 

 structed, gives the following: 



G x = = 1352, £„_<,„ -2680, G x = V4 - 3460 grams. 



This last number indicates the total precipitation falling on a strip 

 one meter wide in one second on the side of the slope that faces the 

 wind. According to the course of the curve 55, as shown in fig. 3, 

 the precipitation begins, first, in the neighborhood of x = — 0.108^ 

 and therefore is distributed along a strip of the ground surface, 

 whose length is 0.358,}, or 8600 meters; from this we compute the 

 average precipitation per hour, as follows: 



3.6 X 3460 , AK . .. 

 = 1 .45 mm. depth 



8600 



= 1.45 kg. mass 



