718 PROCEEDINGS OF THE AMERICAN ACADEMY. 



as the units of length and of time respectively, the differential equation 

 becomes 



9' 2 u 9 2 u 



and the fundamental rectaugle is two adjacent unit squares. 



Such surfaces have been plotted from the data of Table II, for the 

 aliquot cases, from \ to f, and are shown in the top line of the first of 

 the accompanying plates. They have a number of common characteris- 

 tics, and, in particular : 



1. The displacement, u, is zero not only along the lines x = and x = 1, 

 but along the lines t=0, t = \, and t = 2 as well. 



2. Each surface is made up entirely of plane facets, whose projec- 

 tions on the basal x t plane are the squares formed by two sets of 

 equidistant lines, parallel to the diagonals of the large squares, and deter- 

 mined as follows : one line of each set passes through A, the point of 

 discontinuity in the known section, u = 4> (t), under the bow (see Fig- 

 ure 7) ; if the points where these lines intersect the line t = be called 

 B and C, B determines a parallel adjacent to AC, and C one adjacent 

 to AB. 



Conversely, if these two properties be assumed, it is possible to recon- 

 struct the whole surface from the known section u = 4> (t). For the set 

 of squares can be laid out in accordance with the second of the above 

 rules; then the heights of the various corners on some arbitrary scale 

 can be determined as follows (see Figure 7). Assume the height of A 

 to be 4. Then D must be 2, for the known section drops uniformly 

 from A to E. Since A is 4 above C, F must be 4 above G, or 

 + 4 = 4. 18 Then since F is 4 above H, I must be 4 above D, or 

 2 + 4 = 6. And I with A and C determines J. Furthermore, the 

 heights of the three corners still undetermined must be equal to those 

 of D, I, and J respectively, since E = K. The intermediate heights along 

 the sides of the small squares and the contours within them can now be 

 filled in without difficulty. 



Fortunately, this same process, when carried through for any rational 

 case that obeys Krigar-Menzel's law, gives the desired solution graphi- 

 cally. To show how it works in the general case, the construction of 

 the surface for § will be described in some detail with the help of 



18 The drop along one edge of an oblique section of a square prism must be 

 equal to that along the edge opposite, and therefore to that along the opposite side 

 of an adjacent prism, etc. 



