B. A. Keen 411 



At first sight the figures appear to show a remarkable relation. If 

 Y21 has any value greater than about 9, the actual amount of unfree 

 water increases and then decreases, while the free water continually 

 decreases. It is not imtil Fgi is less than 9 that the free and unfree 

 water both decrease together. The results are shown graphically in 

 Fig. 4, where Y„ is plotted against Z„^. It is necessary to understand 

 that the same general equation (13) holds for each curve in Fig. 3. The 

 actual changes in the mimerical values of Z„ , mentioned above, are due 



to the fact that Fji, and therefore -j, have a different value for each 



curve 2. 



It is clear from Fig. 4 that if the upper curves were produced they 

 too would show a maximum value for Z„, but it would eventually 

 correspond to a total moisture content of a greater amount than the 

 undisturbed soil could take up, and is therefore not compatible with 

 the actual conditions. Obviously if Fji ^ere known the true curve 

 showing Z„ and Y„ would also be expressible by equation (13). The 

 question then arises as to which one of the family of curves shown in 

 Fig. 4 best represents the true amounts of free and unfree water in the 

 soil solution, at varying total moisture contents. 



We obtain no help in this respect from a knowledge of the fact 

 pointed out above (p. 407) that the percentage ratio of unfree water to 

 total moisture increases as the latter diminishes, because, on testing, it 

 wiU be seen that each series of values of Z„ in Table V fulfils this con- 

 dition. It would seem probable that Y.^i must be fairly large, in view of 

 the low freezing-point depression at that point, but it is not easy to 



' Actually, Fig. 3 represents a family of parabolas passing through the origin, with their 

 axes inclined to the axes of co-ordinates. 



* A numerical example may make this clearer. Take as a special case of equation (13) : 



z,=atJ-t„. 



and let A, the constant, have the values 100 and 5. By simple calculation we then have, 

 in the two cases: 



