420 THE USE OF YIELD TABLES 



yield table can be constructed by standard methods from selected plots 

 in the groups of which the forest is composed. From this yield table 

 two ages are chosen, representing respectively the 3'ounger and the 

 older age class. The development of the normal stand as indicated 

 by its current and its mean annual growth is the basis for this choice 

 of ages. 



2. The ages thus chosen from the yield table must then be correlated 

 with a given diameter since it is impossible, in the forest, to determine 

 either the age or area of age classes directly. 



This requires a table of diameter growth on the basis of age, for the 

 species and site (§ 267 to § 269) based on a sufficient number of trees 

 to insure a reliable average. Age is the direct basis of this curve, and 

 not diameter (§ 275). From this table, the diameter sought is indicated, 

 for each of the two age classes. 



3. The total volume on the area contained in the two age classes 

 can be separated into the volume in each age class, by means of these 

 two trees of average diameter, representing average age of each class. 

 This requires: 



(a) That the average volume contained in a tree of this average 

 diameter be found. For this purpose, a curve of average height based on 

 diameter is constructed for the site (§ 209). With the height of a tree 

 of the required diameter thus indicated, its volume is found from the 

 standard volume table for the species and region. 



(b) That the number of trees with this average volume be found 

 for each age class, which is required to make up the total volume of the 

 combined group. This number, multiplied by the average volume 

 will give the volume of each age class. 



This solution is simple, when the total number of trees and their total volume 

 are known. Deducting a given number of trees of a given average volume from the 

 group leaves a residual volume, which is equivalent to a fixed number of trees of the 

 average volume for the remaining group; i.e., with total number, total volume, and 

 the average volume of each tree of two groups fixed, there can be but one solution by 

 which the number in each group, and consequently the sum of their volumes equals 

 the required or existing estimate or total in the stand. . 



If x = number of trees in younger group; 

 y = number of trees in older group ; 

 a = volume of average younger tree; 

 6= volume of average older tree. 



Then 

 and 



a;+?/ = total number of trees in stand, c 



ax +by = total volume of stand, d. 

 If all the trees c had the volume a then instead of a total volume d, 

 ax+ay = ac. 



