Outlines of a Mechanical Theory of Storms Part 4

[Ill.u.s.tration: Fig. 8]

Let XX now represent the axis of the lunar orbit, and C the centre of gravity of the earth and moon, X'X' the axis of the vortex, and KCR the inclination of this axis. Then from

similarity Ct : Tt :: Cm : Mm but Tt : Mm :: Moon's ma.s.s : Earth's ma.s.s.

That is Tt : Mm :: TC : MC.

Therefore the system is still balanced; and in no other point but the point C, can the intersection of the axes be made without destroying this balance.

It will be observed by inspecting the figure, that the arc R'K' is greater than the arc RK. That the first increases the arc AR, and the second diminishes that arc. The arc R'K' is a plus correction therefore, and the smaller arc RK a minus correction. If the moon is between her descending and ascending node, (taking now the node on the ecliptic,) the correction is negative, and we take the smaller arc. If the moon is between her ascending and descending node, the correction is positive, and we take the larger arc. If the moon is 90 from the node, the correction is a maximum. If the moon is at the node, the correction is null. In all other positions it is as the sine of the moon's distance from the nodes. We must now find the maximum value of these arcs of correction corresponding to the mean inclination of 2 45'.

To do this we may reduce TC to Tt in the ratio of radius to cosine of the inclination, and taking TS for radius.

[Ill.u.s.tration: Fig. 9]

{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

n ------- a + n (n - ------- ) sin d.

2 k = a - ---------------------- R

If the moon is between her ascending and descending node, then

n ------- 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 pa.s.sing 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:

[Ill.u.s.tration: Fig. 10]

Let the curve pa.s.sing through C represent a portion of the orbit of the earth, and S the sun. From the principles laid down, the density of the ethereal medium increases outward as the square roots of the distances from the sun. Now, if we consider the circle whose centre is C to represent the whole terral vortex, it must be that the medium composing it varies also in density at different distances from the sun, and at the same time is rotating round the centre. That half of the vortex which is exterior to the orbit of the earth, being most dense, has consequently most inertia, and if we conceive the centre of gravity of the earth and moon to be in the orbit (as it must be) at C, there will not be dynamical balance in the terral system, if the centre of the vortex is also found at C. To preserve the equilibrium the centre of the vortex will necessarily come nearer the sun, and thus be found between T and C, T representing the earth, and ? the moon, and C the centre of gravity of the two bodies. If the moon is in opposition, the centre of the vortex will fall between the centre of gravity and the centre of the earth, and have the apparent effect of diminis.h.i.+ng the ma.s.s of the moon.

If, on the other hand, the moon is in conjunction, the centre of the vortex will fall between the centre of gravity and the moon, and have the apparent effect of increasing the ma.s.s of the moon. If the moon is in quadrature, the effect will be null. The coefficient of this inequality is 90', and depends on the sun's distance from the moon. When the moon is more than 90 from the sun, this correction is positive, and when less than 90 from the sun, it is negative. If we call this second correction C, and the moon's distance from her quadratures Q, we have the value of C = (90' sin Q)/R.

[Ill.u.s.tration: Fig. 11]

This correction, however, does not affect the inclination of the axis of the vortex, as will be understood by the subjoined figure. If the moon is in opposition, the axis of the vortex will not pa.s.s through C, but through C', and QQ' will be parallel to KK'. If the moon is in conjunction, the axis will be still parallel to KK', as represented by the dotted line qq'. The correction, therefore, for displacement, is equal to the arc KQ or Kq, and the correct position of the vortex on the surface of the earth at a given time will be at the points Q or q and Q'

or q', considering the earth as a sphere.

[Ill.u.s.tration: Fig. 12]

In the spherical triangle APV, P is the pole of the earth, V the pole of the vortex, A the point of the earth's surface pierced by the radius vector of the moon, AQ is the corrected arc, and PV is the obliquity of the vortex. Now, as the axis of the vortex is parallel to the pole V, and the earth's centre, and the line MA also pa.s.ses through the earth's centre, consequently AQV will all lie in the same great circle, and as PV is known, and PA is equal to the complement of the moon's declination at the time, and the right, ascensions of A and V give the angle P, we have two sides and the included angle to find the rest, PQ being the complement of the lat.i.tude sought.

We will now give an example of the application of these principles.

_Example._[10] Required the lat.i.tude of the central vortex at the time of its meridian pa.s.sage in longitude 88 50' west, July 2d, 1853.

CENTRAL VORTEX ASCENDING.

Greenwich time of pa.s.sage 2d. 3h. 1m.

Mean longitude of moon's node 78 29'

True " " 79 32 Mean inclination of lunar orbit 5 9 True " " 5 13 Obliquity of ecliptic 23 27 32?

Mean inclination of vortex 2 45 0

Then in the spherical triangle PEV,

PE is equal 23 27' 32?

EV " 7 58 0 E " 100 28 0 P " 18 5 7 PV " 26 2 32

Calling P the polar angle and PV the obliquity of vortex.

[Ill.u.s.tration: Fig. 13]

To find the arc AR.

By combining the two proportions already given, we have by logarithms:

M.R.V. minor = 3256 Log. 3.512683 M.S.D. of moon = 940? " 2.973128 P.S.D. of earth = 3950 A. C. 6.403403 Radius 10.000000 T.S.D. of moon 885?.5 A. C. 7.052811 Log. Cosine arc AR = 28 57' 3? 9.942025 ---------

As the only variable quant.i.ty in the above formula is the "True"

semi-diameter of the moon at the time, we may add the Constant logarithm 2.889214 to the arithmetical complement of the logarithm of the true semi-diameter, and we have in two lines the log. cosine of the arc AR.

We must now find the arc RK equal at a maximum to 2 45'. The true longitude of the moon's node being 79 32', and the moon's longitude, per Nautical Almanac, being 58 30', the distance from the node is 21 2', therefore, the correction is

-2 45' sin 21 2'

-arc RK = --------------------- = -59' 13?

R

To find the correction for displacement.

True longitude of sun at date 100 30'

" of moon " 58 30 Moon's distance from quadrature 48 0

As the moon is less than 90 from the sun this correction is also negative, or