272 Rev. 0. Fisher — On Faulting, Jointing, and Cleavage. 



The first question is whether these surfaces will be inclined 

 synclincally or anticlinally towards the crest at K. The horizontal 

 compression is caused by the tendency in the tract as a whole to 

 become flattened by the excess of its weight over the support from 

 below. It is clear, therefore, that the movement by which this 

 flattening is effected, must tend to increase the horizontal stress. 

 The planes of shearing must therefore hade away from those portions 

 of the tract which sink most, as at P, Q, and R in Fig. 3 (below), 

 and if there be any such, towards those which rise relatively to the 

 others, as at places intermediate between P and Q and Q and B. 



Let Cc be a surface on the opposite side of the crest, so situated 

 that the velocity along it is uniform, and equal to that along Bb. If, 

 then, we draw two horizontal lines, BQ, rq, we have the velocities 

 at B, Q, r, q, all equal to one another. But it is obvious that, whereas 

 the whole mass BbcC is tending to pass, relatively to the rest of the 

 tract, through be, the passage being kept from being dilated by the 

 horizontal pressure, the particles along rq must, on the average, be 

 moving downwards faster than those along BQ, because rq is the 

 narrower passage. But at B, Q, r, q, the velocities are all equal. 

 Therefore the velocity must increase faster on going from r towards 

 q than it does on going from B towards Q. If then we seek two 

 points T, t, where the velocity shall be the same, the interval BT 

 will be greater than rt. And points so determined will give another 

 surface of shear and cleavage. We see then that, because BT is 

 greater than 7% the cleavage planes will become more and more 

 nearly vertical as they are less and less distant from the crest. And 

 at the crest, where their inclination changes, they will be vertical. 

 But, the change of velocity close to the crest being small, we may 

 expect the cleavage there to be ill developed. 



We will now appx'oximately estimate the pressures within the 

 tract under consideration. Draw a curve through P, which holds 

 everywhere the same relative position between LKN and HSM : and 

 let, 



KS =h, KP = X, 



w =z the weight of unit mass, 



S = the upwards pressure from below at S ; 



P =: the tangential pressure through P upon unit area of the 

 vertical section, 



r =z the radius of curvature at P. 

 Then the downward force acting on the column KS of unit section 



is wh — fe ; and this will produce an acceleration — r— on all parts 

 of the column. 



Treating every curved slice, such as that through P, independently, 

 it results from a well-known proposition that the tangential stress 

 at P will be, 



P = r. 



h 



' See " Physics of tte Earth's Crust," p. 36, where the calculation in a somewhat 

 similar case is made. 



