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plane of repose, CF that of the plane of fracture, and the ar- 

 row R that of the resistance : put 



c = the angle of repose. 



c = the complement of the angle of repose. 



/3 = the angle DCE contained between the plane of re- 

 pose and the face of the bank. 



5 = the supplement of the sum of the complement of the 

 angle of repose, and the angle which the given direction of the 

 resistance makes with the face. 



6 = the angle KDF, contained between the face produced 

 and the top of the bank. 



^ = the angle DCF, contained between the plane of frac- 

 ture and the face. 



h = the length of the face CD. 



tv = the weight of a cubical unit of the bank. 



R = the resistance. 



Then, when the resistance is a maximum, 



tan /3 V (tan 9 tan g) 



^"^~ v/ (tan 0tang) + V {(tan /3 + tang) X (tan 0- tan j3))' ^^^ 



R = 



wh^ tan 9 sin j3 tan j3 



( 



(2) 



2 cosg 



1 



v/{tanS(tan0-tanj3)} + -v/{tan0 (tan g + tan j3) } 



Equation (1) furnishes the following 'geometrical con- 

 struction for finding the fracture CF. Draw any line GH at 

 right angles to the face produced, cutting the slope DE at H 

 and the line DG ; making the angle GDK = gat G : on GH 

 describe a semicircle cutting the face produced in I : draw De 

 parallel to the plane of repose, CE, meeting GH in e: draw 

 eO parallel toKI, meeting the circumference in O : make IL 

 equal eO ; draw 1/ parallel to Le, and CF parallel to D/; 

 CF is the fracture requiring a maximum resistance to sustain 

 the bank CDF. 



