From the earth nebula 234,620,000 miles in diameter, 6,762,303,076,923,031^{9} cubic miles in volume, and 14,024,781 times less dense than water, we have to subtract the volume of the ring of the earth's system, which, in Table II., appears as 1,489,310,236,000 cubic miles at density of water. Multiplying this by 14,024,781 we find it to have been 20,887,249,553^{9} cubic miles at the same density as the nebula. And subtracting this quantity from 6,762,303,076,923,031^9, we get 6,762,282,189,673,478^9 cubic miles for the volume of the previous nebula after the separation of the ring for the system of the earth. For finding the dimensions of the ring we have 185,930,000 miles for the mean diameter of the Earth's orbit, which makes the circumference 584,117,688 miles in length, and dividing the volume of the ring for the system, which was found to be 20,887,249,553^9 cubic miles, by this length, the area of its cross section comes to be 35,760,344,109 square miles, which divided by the breadth of 37,205,000 miles--that is one-half of the difference between the diameters of the Earth and Venus nebulæ, respectively 234,620,000 and 160,210,000 miles--makes the thickness of the ring to have been 961 miles. But the inner will presently be seen to have been 3·141 times more dense than the outer edge when its separation was completed, so that the average thickness would be 612 miles. VENUS NEBULA. As the volume of the nebula was 6,762,282,189,673,478^9 cubic miles after the separation of the ring for the system of the Earth, we have to condense it into the volume of the Venus nebula, which at 160,210,000 miles in diameter would be 2,153,120,792,079,208^9 cubic miles. Then dividing the larger of these two volumes by the smaller, we find that the density of the Venus nebula had been increased to 3·141 times what that of the Earth nebula was. But we found the density of that nebula to have been 14,024,781 times less than that of water, dividing which by 3·141 makes the Venus nebula to have been 4,465,512 times less dense than water. Dividing this again by 773·395 we find it to have been 5,774 times less dense than air, which would make its absolute temperature to have been 0·04745486°, which corresponds to -273·9525459°. 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.