The purpose of Dr. Mill’s paper is to show a simple yet adequate basis on which to build the superstructure of physical geography. It does not attempt a discussion of the distribution and varieties of vertical relief. The structure of the earth is stated most simply by describing it as an irregular stony ball, covered with an ocean and an envelope of air. If the lithosphere were perfectly smooth and at rest, with the hydrosphere uniformly spread over its surface, the former would have the form of the terrestrial spheroid, and the latter would surround it to a depth of 1.7 miles. The surface of this hypothetical spheroid Dr. Mill calls _mean sphere level_. Of course, in reality the surface of the lithosphere is not perfectly smooth. Parts of it are greatly depressed and parts much elevated, the latter forming the land of the earth. The writer proceeds to calculate the position of mean sphere level, and in the absence of accurate data he uses the careful estimates of Dr. John Murray, which are as follows: Average depths of oceans = 2.36 miles; average height of land = .426 miles; average thickness of hydrosphere surrounding smoothed lithosphere = 1.7 mile; area of land = 55,000,000 square miles; area of oceans = 141,700,000 square miles. Suppose a block of 55,000,000 of square miles area and 1.7 miles deep to be cut out of the smoothed lithosphere and set down on the surface alongside the depression. No change will take place in the surface of the hydrosphere. If the surface of the 141,700,000 square miles of lithosphere were reduced to uniformity, the whole depressed area would lie .66 mile beneath mean sphere level, and the depth of the ocean becomes 2.36 miles. To raise the land to its actual mean level above the hydrosphere surface, a sufficient quantity of matter must be removed from the depressed area and placed on the elevated block. Let _x_ = the thickness of the belt removed and _y_ equal the thickness of the belt when placed on the elevated block. Then _x_ + _y_ is the height of the land above the actual hydrosphere level. From the data given the following equations are easily obtained: _x_ + _y_ - .426 = 0 141.7 _x_ - 55 _y_ = 0 _x_ = .12 and _y_ = .306 in miles. The average height of the land above mean sphere level is thus 1.7 + .306 = 2.006 miles, and the average depth of the depressed portion beneath mean sphere level is .66 + .12 = .78 mile. Dr. Mill divides the earth into the three following divisions: (1) _Abysmal area_, occupying all the depressions beneath the mean surface of the lithosphere, occupying 50 per cent. of the earth’s surface; (2) _Transitional area_, comprising all the regions above mean sphere level covered by the hydrosphere, occupying 22 per cent. of the surface; (3) _Continental area_, all the lithosphere that projects above the hydrosphere, or 28 per cent. of the earth’s surface. J. A. B. FOOTNOTE