_Method as applied by Professor Haughton._—Professor Haughton estimates the mass of the stratified rocks down to the time of the Miocene Tertiary period as being 177,200 feet in thickness, and covering an area equal to that of the sea. The present rate of subaërial denudation he considers to be equal to one foot removed off the surface of the land in 3,090 years. If the proportion of land to water be taken as 52 to 145, it thus requires 8,616 years to deposit one foot of sediment over the bed of the ocean, and consequently this is the rate at which strata are at present being formed. This would give 8,616 × 177,200 = 1,526,750,000 years for the age of the stratified rocks. But he assumes the rate of denudation to have been _ten times_ greater in geological time than at present. This consequently reduces the age of the rocks to 152,675,000 years. By adding one-third for the time which has elapsed since the Miocene Tertiary period he gets 200,000,000 years as a minimum length of geological time.[32] [Footnote 32: _Physical Geography_, p. 94.] The validity of this result rests upon what appear to me to be two very doubtful assumptions. It is assumed in his calculations that the total amount of strata formed during past ages (not the amount presently remaining) was equal to a mass 177,200 feet in thickness, covering the entire area of the ocean. This is certainly doubtful. It may have been as great, for anything that can be proved to the contrary; but we have no evidence that it was so. Certainly there is no evidence that the rate of subaërial denudation during past ages was ever ten times as great as it is now. But how is a length of 200,000,000 years to be reconciled with the age of the sun’s heat? The stratified rocks may be as old as this, but assuredly they are not if gravitation was the only source from which the sun derived his energy. _Method as applied by Mr. Alfred R. Wallace._—Mr. Wallace adopts Professor Haughton’s estimate of 177,200 feet for the maximum thickness of the sedimentary rocks. But, instead of supposing, like Professor Haughton, the products of denudation to be uniformly spread over the entire sea-bottom, he supposes them spread over a belt of merely 30 miles broad, extending along the entire coast-line of the globe, which he estimates at 100,000 miles. This gives an area of 3,000,000 square miles on which the denuded matter of the whole land area of 57,000,000 square miles is deposited. These two areas are to one another as 1 to 19, and thus it follows that deposition goes on 19 times as fast as denudation. The rate of denudation he takes as one foot removed off the surface of the land in 3,000 years, so that the rate of deposition would be about one foot in 158 years, and consequently the time required to deposit the 177,200 feet of rock would be 177,200 × 158 = 27,997,600 years.