THE FORMS OF SOIL WATER 199 



The simplest way of explaining the application of these 

 methods for the expression of the amount of water in a 

 soil is by a specific case. Suppose a certain soil in field 

 condition weighs 100 pounds to a cubic foot and carries 

 10 pounds of water. Obviously it would contain 10 

 per cent of water by the wet method of calculation, or 

 11.1 per cent of water, using the absolutely dry soil as 

 a basis. A pound of water contains 27.6 cubic inches; 

 therefore the amount of water carried by this soil expressed 

 by volume would be 276 cubic inches for every cubic 

 foot of soil. The percentage by volume would equal 

 (276 -f- 172S) X 100, or about 16 per cent. An inch of water 

 covering the top of a cubic foot weighs 5.2 pounds. Ob- 

 viously the number of surface inches which this 10 pounds 

 of water would occupy if placed on the top of the cubic 

 foot of soil would be 10 •*■ 5.2 or 1.92 surface inches. 



The first method of moisture expression, as percentage 

 on a wet basis, is open to two serious objections. In the 

 first place, two different percentages of water in different 

 samples of the same soil do not represent the same degrees 

 of wetness as are expressed by the percentages. For 

 example, 100 grams of wet soil containing 5 per cent of 

 water would consist of 5 grams of water and 95 grams of 

 soil, a ratio of 1 to 19. If the soil contained instead 25 

 per cent of water, the ratio would be 1—3 instead of 1—3.8, 

 as the ratio of the percentages would naturally lead one 

 to expect. The second objection is just as serious and 

 arises from the fact that soils have different apparent 

 weights. For example, 5 per cent of water on the wet 

 basis for a clay weighing when dry 70 pounds to the cubic 

 foot would equal 3.68 pounds, while 5 per cent on a sand 

 weighing 100 pounds would give 5.26 pounds of the same 

 volume. The error of such a method of expression is 



