Jan. 28, 1918 



Study of Plow Bottoms 



159 



To find the geometric center, substitute the coefficients from equation 

 (39) into equations (22), (23), and (24). Solving, we find 



X(,= — 1.405 inches. 

 yo= 6.52 inches. 

 2o= 16.4 inches. 



This translation of axes is shown in figure 9. From equation (25a) 

 G=— 57.3. From (25) the equation of the surface referred to parallel 

 axes through the center is 



3.9XH/+ 3-452"- 7-5372- 7-28«2+ 6.79x^-57.3 = 0. (40) 



y 



Fig. 9. 



To find the equation of the surface referred to the principal axes through 

 the center, substitute the coefficients from (39) into (26), and we have 



fe'-8.35fe2_ 20.17^+2.45 = 0. (41) 



On solving by Horner's method 



ki= 10.27 

 ^2= 0.128 

 ^3= -2.05 



Substituting the values just found for k^, k^, k^, D, and G in equation 



(27), we find 



or 



10.27*^+. 128^^—2.052-= 57. 3 



^ y^ 



I. 



(42) 



(2.36)2 (21.2)2 (5.29)^ 



The direction cosines of the angles which the axes make after rotation 

 with the original axes are obtained by making the proper substitutions in 

 equations (28), (29), (30), and (31). 



