{TC × Cos &c. (inclination 2° 45′)}/R is equal the cosine of the arc SK′ and SK′ + AS = AK′ and AK′ + AR′ = R′K′. But from the nature of the circle, arc RK + arc R′K′ = angle RCK + angle R′CK′, or equal to double the inclination; and therefore, by subtracting either arc from double the inclination, we may get the other arc. The maximum value of these arcs can, however, be found by a simple proportion, by saying; as the arc AR, plus the inclination, is to the inclination, so is the inclination to the difference between them; and therefore, the inclination, plus half the difference, is equal the greater arc, and the inclination, minus half the difference, is equal the lesser; the greater being positive, and the lesser negative. Having found the arc AR, and knowing the moon's distance from either node, we must reduce these values of the arcs RK and R′K′ just found, in the ratio of radius to the sine of that distance, and apply it to the arc AR or A′R′, and we shall get the first correction equal to the arc AK or AK′. Call the arc AR = a " inclination = n " distance from the node = d " arc AK = k and supposing the value of AK be wanted for the northern hemisphere when the moon is between her descending and ascending node, we have n² ------- a + n (n - ------- ) sin d. 2 k = a - ---------------------- R If the moon is between her ascending and descending node, then n² ------- a + n (n - ------- ) sin d. 2 k = a + ---------------------- R The computation will be shorter, however, if we merely reduce the inclination to the sine of the distance from the node for the first correction of the arc AR, if we neglect the semi-monthly motion of the axis; for this last correction diminishes the plus corrections, and the first one increases it. If, therefore, one is neglected, it is better to neglect the other also; especially as it might be deemed affectation to notice trifling inequalities in the present state of the elements of the question. There is one inequality, however, which it will not do to neglect. This arises from the displacement of the axis of the vortex. DISPLACEMENT OF THE AXIS. We have represented the axis of the terral vortex as continually passing through the centre of gravity of the earth and moon. Now, by following out the principles of the theory, we shall see that this cannot be the case, except when the moon is in quadrature with the sun. To explain this: [Illustration: Fig. 10]