[6] The principal cause of these waves is, no doubt, due to the vortices, and the eastern progress of the waves due to the rotating ether; but, at present, it will not be necessary to separate these effects. [7] The inner vortex may reach as high as 83° when the moon's orbit is favorably situated. [8] The curvature of the earth is more than 10 miles in a distance of 300 miles. [9] In middle latitudes. SECTION SECOND. MECHANICAL ACTION OF THE MOON. We will now proceed to give the method of determining the latitude of the axis of the vortex, at the time of its passage over any given meridian, and at any given time. And afterwards we will give a brief abstract from the record of the weather, for one sidereal period of the moon, in order to compare the theory with observation. [Illustration: Fig. 4] In the above figure, the circle PER represents the earth, E the equator, PP′ the poles, T the centre of the earth, C the mechanical centre of the terral vortex, M the moon, XX′ the axis of the vortex, and A the point where the radius vector of the moon pierces the surface of the earth. If we consider the axis of the vortex to be the axis of equilibrium in the system, it is evident that TC will be to CM, as the mass of the moon to the mass of the earth. Now, if we take these masses respectively as 1 to 72.3, and the moon's mean distance at 238,650 miles, the mean value of TC is equal to this number, divided by the sum of these masses,--_i.e._ the mean radius vector of the little orbit, described by the earth's centre around the centre of gravity of the earth and moon, is equal 238650/(72.3+1) = 3,256 miles; and at any other distance of the moon, is equal to that distance, divided by the same sum. Therefore, by taking CT in the inverse ratio of the mean semi-diameter of the moon to the true semi-diameter, we shall have the value of CT at that time. But TA is to TC as radius to the cosine of the arc AR, and RR′ are the points on the earth's surface pierced by the axis of the vortex, supposing this axis coincident with the pole of the lunar orbit. If this were so, the calculation would be very short and simple; and it will, perhaps, facilitate the investigation, by considering, for the present, that the two axes do coincide. In order, also, to simplify the question, we will consider the earth a perfect sphere, having a diameter of 7,900 miles, equal to the actual polar diameter, and therefore TA is equal to 3,950 miles. In the spherical triangle given on next page, we have given the point A, being the position of the moon in right ascension and declination in the heavens, and considered as terrestrial latitude and longitude.