Penny Blood: Serial Sensation and Working-Class Entertainment — page 125: what you’re looking at

Machine-transcribed from the original scan — historical spelling and the odd misread are preserved.

108 CRUST NOT FLEXIBLE. [cH. 1x. But none but integral values of m will be compatible with equilibrium, because the curve must meet either end of the trough at an anticlinal. ¢@ diminishes as m increases, and when m is infinitely great and ¢ infinitely small, | sin p as =]l—e p 2 becomes unity, and therefore e = 0, or there is no compression. m also increases as & diminishes. Hence, ceteris paribus, the festoons are more numerous with a thin crust than with a thick one. The geometrical relations show that the lengths of the festoons must increase, and their number diminish, as the compression is increased. Consider now the crust to be in equilibrium with a certain pair of values of m and e, m being integral, and let the trough then be shortened, but not sufficiently so for the next integral value of m to be reached. What will happen? The festoons cannot alter their curvature so as to adapt themselves to their lessened chords. It seems then that the compression at the anticlinals must become finite, instead of being zero, and that the ends of the festoons will be pushed up against each other, by which means material will be accumulated there, and Q and £ will become finite at the anticlinals. This will accord with the undulating form of the curve, but it seems that the equilibrium must become unstable, on account of the weight resting on the anticlinal in the highest possible position. It will therefore be liable to slide down upon the surface of the liquid on one side or other of the anticlinal. Let us suppose then that a portion of the crust PQ, whose weight is Q, has become thickened by the accumulation of material, and let P be the mean fluid-pressure upon this portion of crust. And let it be supposed that the curve is cut off above Q. . 7 COMIC OOOKS.EO