From the Venus nebula of 160,210,000 miles in diameter, volume 2,153,120,792,079,207,921^{ 6} cubic miles, and density 4,465,512 times less than that of water, we have now to deduct the volume of her ring, which by Table II. is 1,131,960,000,000 cubic miles at the density of water. Multiplying this volume by 4,465,512 we find the volume of the ring to have been 5,054,780,604,651^{ 6} cubic miles at the same density as the nebula, and subtracting this amount from 2,153,120,792,079,207,921^{ 6} we get 2,153,115,737,298,603^{ 6} cubic miles for the volume to be condensed into the nebula following.
To find the dimensions of the ring we have 134,490,000 miles for the diameter of the orbit of Venus, which makes its circumference 422,513,784 miles in length. Then dividing the volume of the ring, i.e. 5,054,780,604,651^{ 6} cubic miles by this length, the area of its cross-section comes to be 11,963,821,788 square miles, which, divided by the breadth of 28,489,000 miles--that is one-half of the difference between the diameters of the Venus and Mercurian nebulæ, respectively 160,210,000 and 103,232,000 miles--makes the thickness of the ring to have been 420 miles. But the inner edge having become, in the process of separation, 3·738 times more dense than the outer one (see below) the average thickness would be reduced to 225 miles.
As the volume of the nebula was 2,153,115,737,298,603,270^{ 6} cubic miles after the separation of the ring for Venus, we have to condense it into the volume of the Mercurian nebula, which at 103,232,000 miles in diameter would be 576,026,613,333,333,333^{ 6} cubic miles. Then, dividing the larger of these two volumes by the smaller, we find that the density of the Mercurian nebula must have been increased 3·738 fold over that of its predecessor. But we find the density of the Venus nebula to have been 4,465,512 times less than water, dividing which by 3·738 makes the Mercurian nebula to have been 1,194,666 times less dense than water. Dividing again this density by 773·395 we find it to have been 1545 times less than air, and 274° divided by this air density gives 0·1773463° as its absolute temperature, which corresponds to -273·8226537°.
From the Mercurian nebula 103,232,000 miles in diameter, volume of 576,026,613,333,333,333^{ 6} cubic miles, and density of 1,194,666 times less than water, we have to deduct the volume of his ring, which by Table II. is 92,735,000,000 cubic miles at density of water. Multiplying this volume by 1,194,666 makes the ring to have been 110,787,355,300^{ 6} cubic miles in volume at the density of the nebula, and subtracting this amount from 576,026,613,333,333,333^{ 6}, we get 576,026,502,545,978,033^{ 6} cubic miles for the volume to be condensed into the nebula following.