Thus by a shock in the direction O C, a ball, A, on the top of a pedestal would describe a trajectory to the point C. Let the angle which O C makes with the horizon be _e_, the vertical height through which the ball has fallen be _b_, and the horizontal distance of projection be _a_; then _a_^2 _b_ = _a_ tan _e_ + ————————————, 4H cos^2 _e_ H being the height due to the velocity of projection. Whence ___________________ 2H ± √4H(H + _b_) - _a_^2 Tan _e_ = ——————————————————————————. _a_ _a_^2 _g_ V^2 = ———————————————————————————————. 2 cos^2 _e_ (_b_ - _a_ tan _e_) For the back motion or subnormal wave in the direction C O, ___________________ 2H ± √4H(H + _b_) - _a_^2 Tan _e_ = ——————————————————————————. _a_ _a_^2 _g_ V^2 = ———————————————————————————————. 2 cos^2 _e_ (_b_ + _a_ tan _e_) A serious error which may enter into calculations of this description when practically applied has been pointed out when speaking of columns as seismometers. It was then shown that such bodies before being overthrown may often be caused to rock, and therefore that their final overthrow may not have any direct connection with the impulse of the succeeding shock. Another point to which attention must be drawn respecting the above calculations is that if there was no friction or adherence between the projected body and its pedestal, in consequence of its inertia it would be left behind by the forward motion of the shock, and simply drop at the foot of its support. In the case of frictional adherence it would be carried forward by the velocity acquired before this adherence was broken, and thrown in a direction _opposite_ to that given in the figure—that is to say, in the direction of the shock.[14]